Introductory Electrical Engineering With Math Explained In Accessible Language 1119580188, 9781119580188, 1119580226, 9781119580225, 111958020X, 9781119580201

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Introductory Electrical Engineering with Math Explained in Accessible Language

Introductory Electrical Engineering with Math Explained in Accessible Language Magno Urbano Independent Author Lisbon, Portugal

This edition first published 2020 © 2020 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, except as permitted by law. Advice on how to obtain permission to reuse material from this title is available at http://www.wiley.com/go/permissions. The right of Magno Urbano to be identified as the author of this work has been asserted in accordance with law. Registered Office John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, USA Editorial Office 111 River Street, Hoboken, NJ 07030, USA For details of our global editorial offices, customer services, and more information about Wiley products visit us at www.wiley.com. Wiley also publishes its books in a variety of electronic formats and by print-on-demand. Some content that appears in standard print versions of this book may not be available in other formats. Limit of Liability/Disclaimer of Warranty While the publisher and authors have used their best efforts in preparing this work, they make no representations or warranties with respect to the accuracy or completeness of the contents of this work and specifically disclaim all warranties, including without limitation any implied warranties of merchantability or fitness for a particular purpose. No warranty may be created or extended by sales representatives, written sales materials or promotional statements for this work. The fact that an organization, website, or product is referred to in this work as a citation and/or potential source of further information does not mean that the publisher and authors endorse the information or services the organization, website, or product may provide or recommendations it may make. This work is sold with the understanding that the publisher is not engaged in rendering professional services. The advice and strategies contained herein may not be suitable for your situation. You should consult with a specialist where appropriate. Further, readers should be aware that websites listed in this work may have changed or disappeared between when this work was written and when it is read. Neither the publisher nor authors shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. Library of Congress Cataloging-in-Publication Data Names: Urbano, Magno, author. Title: Introductory electrical engineering with math explained in accessible language / Magno Urbano. Description: Hoboken, NJ : Wiley, 2020. | Includes index. Identifiers: LCCN 2019030445 (print) | LCCN 2019030446 (ebook) | ISBN 9781119580188 (paperback) | ISBN 9781119580225 (adobe pdf) | ISBN 9781119580201 (epub) Subjects: LCSH: Electrical engineering–Mathematics. Classification: LCC TK153 .U73 2020 (print) | LCC TK153 (ebook) | DDC 621.3–dc23 LC record available at https://lccn.loc.gov/2019030445 LC ebook record available at https://lccn.loc.gov/2019030446 Cover design by: Wiley Cover image: © Andrew Brookes/Getty Images Set in 10/12pt Warnock by SPi Global, Pondicherry, India Printed in the United States of America 10 9 8 7 6 5 4 3 2 1

v

Contents About the Author xix Preface xxi Acknowledgement xxiii Introduction xxv Conventions: Used by this Book

xxvii

1

Scientific Method: General Concepts

1.1 1.2 1.3 1.4 1.4.1 1.4.2 1.4.3 1.4.3.1 1.4.3.2 1.4.3.3 1.4.3.4 1.4.3.5 1.4.3.6 1.4.4 1.4.5

Introduction 1 Powers of 10 1 Roots 2 Scientific Notation as a Tool 2 Very Large Numbers 2 Very Small Numbers 3 Operations with Powers of 10 3 Multiplication 3 Division 4 Adding and Subtracting 4 Raising to a Number 4 Expressing Roots as Exponents 5 Extracting the Root 5 Computers and Programming 6 Engineering Notation 6

1

2

Infinitesimal Calculus: A Brief Introduction

2.1 2.2 2.2.1 2.2.2 2.2.3

Introduction 9 The Concept Behind Calculus Limits 10 Derivatives 11 Integral 15

3

Atom: Quarks, Protons, and Electrons

3.1 3.2

Introduction 19 Atoms and Quarks

19

9

19

9

vi

Contents

3.3 3.4 3.5 3.6 3.7 3.8 3.9

Electrons 20 Strong Force and Weak Force 21 Conductors and Electricity 22 The Shells 23 Electric Potential 24 Current 25 Electric Resistance 25

4

Voltage and Current: Direct and Alternating Current and Voltage 27

4.1 4.2 4.3 4.3.1 4.3.2 4.3.3 4.4 4.4.1 4.4.2 4.4.3 4.5 4.5.1 4.6 4.6.1 4.7 4.8 4.8.1 4.8.2 4.8.3 4.8.4 4.8.5

Introduction 27 Terminology 27 Batteries 27 Battery Life 28 Batteries in Series 29 Batteries in Parallel 29 Danger Will Robison, Danger! 30 Never Invert Polarities 30 Never Use Different Batteries 30 Short-Circuiting Batteries 30 Direct Current 31 DC Characteristics 31 Relative Voltages 31 Mountains 32 Ground 33 Alternating Current 34 AC Characteristics 34 AC Cycles 34 Period and Frequency 35 Peak-to-Peak Voltage 36 DC Offset 37 Exercises 38 Solutions 39

5

Resistors: The Most Fundamental Component

5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9

Introduction 41 Resistor 41 Electric Resistance 41 Symbols 41 Types of Resistor 42 Power 42 Color Code 42 Potentiometer 44 Trimpots 44

41

Contents

5.10 5.11 5.11.1 5.12 5.13 5.14 5.15

Practical Usage 45 Electric Characteristics 45 Important Facts 45 Resistors in Series 45 Resistors in Parallel 46 DC and AC Analysis 46 Input and Output Synchronism Exercises 48 Solutions 48

6

Ohm’s Laws: Circuit Analysis

6.1 6.2 6.3 6.4 6.5 6.5.1 6.5.1.1 6.5.2 6.5.2.1 6.5.2.2 6.5.3 6.5.3.1 6.5.4 6.5.4.1 6.5.4.2 6.5.5 6.5.5.1

47

51 Introduction 51 Basic Rules of Electricity 51 First Ohm’s Law 52 Second Ohm’s Law 53 Examples 53 Example 1 53 Solution 53 Example 2 53 Solution 54 Another Method to Obtain the Same Result Example 3 55 Solution 56 Example 4 56 Solution 56 Another Method to Obtain the Same Result Example 5 57 Solution 57 Exercises 58 Solutions 59

7

Delta–Wye Conversions: Circuit Analysis 63

7.1 7.2 7.3 7.4 7.5 7.5.1 7.5.1.1 7.5.2 7.5.2.1

Introduction 63 Delta Circuit 63 Delta–Wye Conversion Wye–Delta Conversion Examples 65 Example 1 65 Solution 65 Example 2 67 Solution 67 Exercises 69 Solutions 69

63 65

55

57

vii

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Contents

8

Capacitors: And Electric Charges 73

8.1 8.2 8.3 8.3.1 8.3.2 8.4 8.5 8.6 8.7 8.8 8.8.1 8.8.2 8.9 8.9.1 8.9.1.1 8.9.1.2 8.9.1.3 8.9.1.4 8.9.2 8.9.2.1 8.10 8.10.1 8.11 8.12 8.12.1 8.13 8.14 8.15 8.16 8.17 8.17.1 8.18

Introduction 73 History 73 How It Works 73 Dielectric 75 Construction Methods 76 Electric Characteristics 77 Electric Field 78 Capacitance 78 Stored Energy 79 Voltage and Current 81 Current on a Charging Capacitor 81 Voltage on a Charging Capacitor 82 Examples 84 Example 1 84 Solution 84 Before the Pulse 85 During the Pulse 85 After the Pulse 86 Example 2 86 Solution 86 AC Analysis 87 Pool Effect 87 Capacitive Reactance 88 Phase 88 Mathematical Proof 89 Electrolytic Capacitor 91 Variable Capacitors 93 Capacitors in Series 93 Capacitors in Parallel 94 Capacitor Color Code 95 The Code 95 Capacitor Markings 96 Exercises 98 Solutions 98

9

Electromagnetism: And the World Revolution

9.1 9.2 9.3 9.4 9.5 9.6 9.7

Introduction 103 The Theory 103 Hans Christian Ørsted 103 The Right-Hand Rule 105 Faraday First Experiment 105 Faraday Second Experiment 106 Conclusion 107

103

Contents

109

10

Inductors: Temperamental Devices

10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.8.1 10.8.2 10.9 10.10 10.11 10.11.1 10.12

Introduction 109 The Inductor 109 Coils and Magnets 110 Inductance 111 Variable Inductor 111 Series Inductance 112 Parallel Inductance 112 DC Analysis 113 Energizing 114 De-energizing 115 Electromotive Force 116 Current Across an Inductor 116 AC Analysis 116 Alternating Voltage and Current 117 Out of Sync 119 Exercises 120 Solutions 120

11

Transformers: Not the Movie

11.1 11.1.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9 11.10 11.11

123 Introduction 123 Transformers 123 Connected by the Magnetic Field 124 Faraday’s Law 124 Primary and Secondary 124 Real-Life Transformer 125 Multiple Secondaries 125 Center Tap 126 Law of Conservation of Energy 127 Leakage Flux 127 Internal Resistance 128 Direct Current 128 129

12

Generators: And Motors

12.1 12.2 12.2.1 12.3 12.3.1 12.3.2

Introduction 129 Electric Generators 129 How It Works 129 Electric Motor 131 DC Motors 131 AC Motors 132

13

Semiconductors: And Their Junctions

13.1

Introduction

133

133

ix

x

Contents

13.2 13.3 13.3.1 13.3.1.1 13.3.1.2 13.3.1.3 13.3.1.4 13.3.1.5 13.3.1.6 13.3.1.7 13.3.1.8 13.3.1.9

It All Started with a Light Bulb 133 Semiconductors 135 Bipolar Junction 135 Making the Structure More Negative 136 Making the Structure More Positive 137 Pure Magic 137 Reverse Biasing 138 Forward Biasing 138 Biasing Curve 138 Thermal Voltage 139 Barrier Voltage 140 Relation Between Current and Voltage 141

14

Diodes and Transistors: Active Components

14.1 14.2 14.3 14.4 14.5 14.5.1

Introduction 143 Diodes 143 NPN Junction 143 Biasing 144 The Transistor, Finally! 144 How Transistors Work? 145

15

Voltage and Current Sources: Circuit Analysis

15.1 15.2 15.3 15.4 15.5 15.6

Introduction 147 Independent DC Voltage Sources 147 Independent AC Voltage Sources 147 Dependent Voltage Sources 148 Independent Current Sources 149 Dependent Current Sources 149

16

Source Transformations: Circuit Analysis

16.1 16.2 16.3 16.3.1

Introduction 151 The Technique 151 Example 153 Solution 153 Exercises 160 Solutions 161

17

Impedance and Phase: Circuit Analysis

17.1 17.2 17.3 17.3.1

Introduction 165 This Is Just a Phase 165 Impedance 166 Series Impedance 166

143

151

165

147

Contents

17.3.2 17.4 17.5 17.6 17.6.1 17.6.1.1 17.6.2 17.6.2.1 17.7 17.7.1 17.7.1.1 17.7.1.2 17.7.1.3

Parallel Impedances 166 Capacitive Impedance 167 Inductive Impedance 169 Examples 169 Example 1 169 Solution 169 Example 2 171 Solution 171 The Importance of Impedances in Real Life Example 173 Solution 173 What About a Bigger Load? 175 Conclusion 176 Exercises 177 Solutions 177

18

Power: And Work

18.1 18.2 18.3 18.3.1 18.4 18.4.1 18.5 18.5.1 18.5.1.1 18.5.1.2 18.5.2 18.5.2.1 18.5.3 18.5.4 18.6 18.6.1 18.6.1.1

181 Introduction 181 Electric Power and Work 181 Powers in Parallel 182 Conclusion 183 Powers in Series 183 Conclusion 184 “Alternating” Power 184 Two Ovens 184 The First Oven 185 Second Oven 185 The Average Value 185 Average Value of a Sinusoidal Function 185 RMS Value 186 Back to the Second Oven 187 Real, Apparent, and Reactive Power 188 Reactive Power 189 Power Factor 190 Exercises 191 Solutions 192

19

Kirchhoff’s Laws: Circuit Analysis 197

19.1 19.2 19.2.1

Introduction 197 Kirchhoff’s Laws 197 Nodes or Junctions 197

173

xi

xii

Contents

19.2.2 19.2.3 19.2.4 19.3 19.3.1 19.3.1.1 19.3.1.2 19.3.2 19.3.2.1

Mesh 198 Kirchhoff’s First Law 198 Kirchhoff’s Second Law 199 Examples 199 Example 1 199 Solution 200 Meshes 200 Example 2 205 The Analysis 205 Exercises 210 Solutions 211

20

Nodal Analysis: Circuit Analysis 215

20.1 20.2 20.2.1 20.2.1.1 20.2.1.2 20.2.1.3 20.2.2 20.2.2.1 20.2.2.2 20.2.3 20.2.3.1 20.2.3.2

Introduction 215 Examples 215 Example 1 215 Solution 215 Circuit Analysis 216 Kirchhoff’s Laws 216 Example 2 219 Solution 219 Applying Kirchhoff’s Laws Example 3 221 Node A 222 Node B 223 Exercises 226 Solutions 227

21

Thévenin’s Theorem: Circuit Analysis 235

21.1 21.2 21.2.1 21.2.2 21.2.3 21.2.3.1 21.2.3.2 21.2.3.3 21.2.3.4 21.2.3.5 21.2.3.6 21.2.3.7 21.2.3.8

Introduction 235 The Theorem 235 The Equivalent Thévenin Circuit 236 Methodology 236 Example 236 Thévenin Equivalent Resistance 237 Thévenin Equivalent Voltage 239 Node A 240 Node B 241 Node C 242 The Thévenin Voltage 243 Same Behavior 245 Node C 247 Exercises 250 Solutions 251

219

Contents

22

Norton’ Theorem: Circuit Analysis 257

22.1 22.2 22.2.1 22.2.2 22.2.3 22.2.3.1

Introduction 257 Norton’s Theorem 257 Finding Norton Equivalent Circuit 258 Methodology 258 Example 258 Finding the Norton Current Source 258 Exercises 263 Solutions 264

23

Superposition Theorem: Circuit Analysis 269

23.1 23.2 23.3 23.4 23.4.1 23.4.1.1 23.4.1.2 23.4.1.3 23.4.2 23.4.2.1 23.4.2.2 23.4.2.3 23.4.3 23.4.3.1 23.4.3.2 23.4.3.3

Introduction 269 The Theorem 269 Methodology 269 Example 270 First Circuit 270 Node C 271 Relation Between Voltage A and C Voltage B 273 Second Circuit 274 Node A 274 Node C 276 Final Result 278 Checking 278 Node A 279 Node C 279 Voltage Across A and B 280 Exercises 281 Solutions 282

24

Millman’s Theorem: Circuit Analysis

24.1 24.2 24.2.1 24.2.1.1 24.3 24.3.1 24.3.1.1 24.3.2

Introduction 287 Millman’s Theorem The Theory 287 Admittance 290 Examples 291 Example 1 291 Solution 291 Example 2 294 Exercises 295 Solutions 295

287

271

287

xiii

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Contents

25

RC Circuits: Voltage and Current Analysis in Circuits Containing Resistors and Capacitors in Series 297

25.1 25.2 25.2.1 25.2.2 25.2.3 25.3 25.3.1.1 25.3.2 25.3.3 25.3.3.1 25.3.3.2 25.3.4 25.4 25.4.1 25.4.1.1 25.4.1.2 25.4.2 25.4.2.1 25.4.2.2 25.4.2.3 25.4.2.4 25.4.3 25.4.3.1 25.4.4 25.4.4.1 25.4.4.2 25.4.4.3 25.4.4.4

Introduction 297 Charging a Capacitor 297 Charging Voltage 298 Charge Equation 301 Charging Current 305 RC Time Constant 308 Transient and Steady States 308 General Formula 309 Discharging a Capacitor 310 Charge During Discharge 311 Voltage During Discharge 314 Current During Discharge 314 Examples 315 Example 1 315 Current Flowing as the Switch Is Turned On 316 Current and Voltage After 1 s 316 Example 2 317 Current Before the Switch Is Moved 318 Current as the Switch Is Moved 318 Voltage After 2 s 318 Current After 2 s 319 Example 3 320 After the Switch Is Closed 320 Example 4 322 Circuit’s Impedance 323 RMS Current 325 Power Factor and Phase Angle 325 The Apparent, the Real, and the Reactive Power 326 Exercises 328 Solutions 330

26

RL Circuits: Voltage and Current Analysis in Circuits Containing Resistors and Inductors in Series 341

26.1 26.2 26.2.1 26.2.2 26.3 26.3.1 26.3.2 26.3.2.1

Introduction 341 Energizing 341 Current During Energizing 342 Voltage During Energizing 346 De-energizing 349 Current During De-energizing 349 Voltage During De-energizing 352 RL Time Constant 353

Contents

26.3.2.2 26.4 26.4.1 26.4.1.1 26.4.1.2 26.4.2 26.4.2.1 26.4.2.2 26.4.2.3 26.4.2.4 26.4.3 26.4.3.1 26.4.3.2 26.4.3.3 26.4.3.4

Transient and Steady States 353 Examples 354 Example 1 354 Current After 10 ms 355 Final Current 355 Example 2 355 Current Before the Switch Is Moved 356 Current as the Switch Is Moved 356 Voltage After 5 ms 357 Current After 5 ms 357 Example 3 358 The Circuit’s Impedance 358 RMS Current 359 Power Factor 360 The Apparent, the Real, and the Reactive Power Exercises 362 Solutions 365

27

RLC Circuits: Part 1: Voltage Analysis in Circuits Containing Resistors, Capacitors, and Inductors in Series 377

27.1 27.2 27.2.1 27.2.2 27.2.3 27.2.3.1 27.2.4 27.2.4.1 27.2.4.2 27.2.4.3 27.2.4.4 27.3 27.3.1 27.3.1.1

361

Introduction 377 A Basic RLC Series Circuit 377 Circuit Analysis 380 Voltage Across the Capacitor 381 Back to the Equation 383 Steady-State Solution 383 Transient Solution 384 The Roots of the Equation 386 Critically Damped Solution 388 Overdamped Solution 397 Underdamped Solution 401 Examples 408 Example 1 408 What Is the Voltage Across the Capacitor After the Switch Is Closed? 409 27.3.1.2 What Is the Voltage Across the Capacitor Two Seconds After the Switch Is Closed? 409 27.3.1.3 What Is the Circuit’s Natural Resonance Frequency? 412 27.3.2 Example 2 412 27.3.2.1 The Complete Equation for the Voltage Across the Capacitor 413 Exercises 418 Solutions 419

xv

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Contents

28

RLC Circuits: Part 2: Current Analysis in Circuits Containing Resistors, Capacitors, and Inductors in Series 427

28.1 28.2 28.2.1 28.3 28.3.1 28.3.1.1 28.3.2 28.3.2.1 28.3.3 28.3.3.1 28.4 28.4.1 28.4.1.1

Introduction 427 The Circuit 427 Current 428 Current Equations 430 Critically Damped Solution 430 Current Curve 430 Overdamped Solution 431 Current Curve 431 Underdamped Solution 431 Current Curve 432 Examples 432 Example 1 432 What Is the Current Across the Circuit After the Switch Is Closed? 433 28.4.1.2 What Is the Current Across the Circuit Two Seconds After the Switch Is Closed? 433 28.4.2 Example 2 437 28.4.2.1 Solution 438 Exercises 442 Solutions 443 29

Transistor Amplifiers: The Magic Component

29.1 29.2 29.3 29.4 29.5 29.6 29.7 29.8 29.8.1 29.8.1.1

Introduction 451 Transistor as Amplifiers 451 The Water Storage Tank 451 Current Gain 452 Power Supply Rails 452 Amplifying 452 Quiescent Operating Point 453 Amplifier Classes 454 Class A 454 Common Emitter 455 Exercises 477 Solutions 479

30

Operational Amplifiers: A Brief Introduction

30.1 30.2 30.3 30.4 30.5

Introduction 485 Operational Amplifiers 485 How Op-Amp Works 486 Op-Amp Characteristics 488 Typical Configurations 488

451

485

Contents

30.5.1 30.5.2 30.5.3 30.5.4 30.5.4.1 30.5.4.2 30.5.5 30.5.6 30.5.7

Inverting Op-Amp 488 Non-inverting Op-Amp 491 Voltage Follower 493 Non-inverting Summing Amplifier 497 Summing Two Inputs 500 Summing Three Inputs 501 Inverting Summing Amplifier 502 Integrator 505 Differentiator 507

31

Instrumentation and Bench: A Brief Introduction

31.1 31.2 31.2.1 31.3 31.4 31.5 31.6 31.7 31.8 31.8.1 31.9 31.10 31.11 31.12 31.13

Introduction 509 Multimeter 509 True RMS Multimeter 510 Voltmeter 510 Ammeter 511 Ohmmeter 512 Oscilloscope 513 Breadboards 513 Wire Diameter 515 Types of Wires 516 Power Supply 516 Soldering Station 517 Soldering Fume Extractors 517 Lead-Free Solder 517 A Few Images of Real Products 518

Appendix Appendix Appendix Appendix Appendix Appendix Appendix Appendix Appendix Appendix Appendix

A: B: C: D: E: F: G: H: I: J: K:

Index

563

509

International System of Units (SI) 521 Color Code: Resistors 523 Root Mean Square (RMS) Value 525 Complex Numbers 529 Table of Integrals 537 AWG Versus Metric System: Wire Cross Sections Resistors: Commercial Values 541 Capacitors: Commercial Values 543 Inductors: Commercial Values 549 Simulation Tools 557 Glossary 559

539

xvii

xix

About the Author With more than 25 years of experience in the field, Magno Urbano started his professional life by graduating as an Electrical Engineer from USU (Brazil) and worked in fields like computer graphics, visual effects, and programming for two of the largest broadcast companies in two continents: Globo TV Network in Brazil and Radio and Television of Portugal (RTP). Prior to this volume, Magno has written 16 books on multimedia themes and authored nearly 100 articles and 50 multimedia courses for the most important magazines in Europe and South America. Magno has been developing apps for Apple devices since 2008. During this period, he has developed and published about 120 applications for iPhone, iPad, macOS, and Apple TV, some of them hitting 1 or placing in the top 10 in multiple countries for several weeks. Some of his apps can be seen at www.katkay.com. Right now, he is excited to deliver his first book published by Wiley & Sons! Magno can be reached at [email protected].

xxi

Preface Magno Urbano, a professional with an amazing educational background in electrical engineering and an autodidact in programming, is always trying to learn the most he can. He worked for the Visual Effects Post Production Center of Globo Television in Brazil, revealing himself as a great professional. He decided to move to Portugal in 2000 to expand his possibilities. Working as an entrepreneur in Portugal, he created a lot of multimedia courses, published a dozen books, and created hundreds of apps for Apple devices. In everything he does, he always shows competence, know-how, qualification, and knowledge. The good persons are always those who make the difference. Taulio Mello Visual Effects Artist, former VFX specialist for TV Globo

I know Magno Urbano for two decades. In the areas that he has worked, like in multimedia, visual effects, Internet, and software development and in the publishing world, I have always recognized in Magno Urbano a huge effort and a notable dedication. The enthusiasm and quality he applies to his works always lead to something that, without doubt, will be of great interest to those who like this branch of universal culture. Fernando Salgado Former Responsible for the Post-Production Center of RTP (Radio and Television of Portugal)

xxiii

Acknowledgement Thanks to my mother and sister and to people inside Wiley & Sons, specially Brett Kurzman, Victoria Bradshaw, Blesy Regulas, and Grace Paulin who made this project possible.

xxv

Introduction 1

Knowledge is Everything

We live in a world where technology is advancing at an astonishing speed. We have arrived at this point in knowledge, because man always had this curiosity about the things around. At first, it was a matter of survival, but then curiosity turned into discovery, and we have started to unveil the secrets of the universe. The human race produced geniuses like Isaac Newton, Copernicus, Galileo Galilei, Alessandro Volta, Michael Faraday, Albert Einstein, Pierre-Simon Laplace, Carl Friedrich Gauss, Richard Feynman, and so many others, who created the pillars of knowledge and contributed to the various branches of science, including physics and mathematics, and in fields that are pertinent to this book, such as electromagnetism and electricity. In this book, we try to present knowledge in a simple and straightforward way and in an accessible language, using as many illustrations and examples as possible. Unlike other similar books, we provide the mathematical theory necessary to the reader, so every theme can be fully understood. Many electrical engineering teachers complain that students arrive at their courses without the previous mathematical skills required to understand the theories. This book tries to fill this gap, by providing such mathematical skills every time they are necessary. We invite the reader to this journey.

xxvii

Conventions Used by this Book

1

Introduction

In this chapter, we will examine the several conventions used in this book.

2

Equations

In mathematics, several conventions are used to show the product or multiplications of two elements like A×B A B AB A B Every one of these conventions represents value A multiplied by value B. In this book we will use all these forms, the one that makes more sense for the context being explained.

3 3.1

Electric Schematics Connection

Every time a small black circuit is drawn in the intersection between two elements, it means that these elements are electrically connected.

xxviii

Conventions: Used by this Book

Figure 0.1 Connected components.

In Figure 0.1, resistor R3 is electrically connected to resistor R2 in one side and to RE on the other. 3.2 Unspecified Value If a component does not have its value specified in a schematic in this book, the following rules must be followed: every resistor is always expressed in Ohms, every capacitor in Farads, every inductor in Henries, and every power supply in Volts. All resistors in Figure 0.1, for example, have their values in Ohms.

4

Mathematical Concepts

Unlike other works, this book provides the reader with the mathematical concepts necessary for understanding a given segment or the development of an equation. In the following example, we see an equation followed by a mathematical concept. The equation following the concept is a result of the concept applied: du = d V1 − d

Q C

MATHEMATICAL CONCEPT The derivative of a constant is 0. 0

du = d V1 − d

Q C

Conventions: Used by this Book

5

Examples and Exercises

This book contains several examples and exercises. The examples are used to illustrate the theory and are resolved promptly. The exercises should be done by the readers. The endings of the chapters contain the answers to the exercises.

xxix

1

1 Scientific Method General Concepts

1.1

Introduction

In this chapter, we will examine the scientific method and its concepts. These concepts must be well understood and are fundamental to achieve a solid foundation in electrical engineering.

1.2

Powers of 10

In science it is common to represent numbers like 100, 1000, and 10000, as 102, 103, and 104, respectively, in what is called powers of ten. Powers of 10 show how many times 10 has to be multiplied by itself to obtain the final value. In other words, 104 = 10 × 10 × 10 × 10 is equal to 10 multiplied by itself 4 times. However, if we want to represent very small numbers like 0.1, 0.01, and 0.001, we have to use negative exponents, as 10−1, 10−2, 10−3, respectively. This negative exponent represents how many times the number must be divided by 10. Numbers like 10−1, 10−2, 10−3 represent 1/10, 1/100, 1/100, respectively.

Any number raised to 0 is equal to 1. Thus, 100 = 1.

Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

2

1 Scientific Method

1.3 Roots Roots are the inverse operation of powers. Roots are expressed in scientific notation as a fractional exponent: 1

2

5 = 52

3

5 = 53

4

5 = 54

4

1

1

2

52 = 54 m

Any number raised to a fractional exponent in the form x n is equal to

n

xm.

1.4 Scientific Notation as a Tool In science, we constantly have to deal with numbers that are very hard to write in natural form, for example, 0.000000000003434033323. Imagine how hard it would be to compare this number with 0.0000000000000212817716 and know which one is bigger. To solve this problem and make it easy to compare and write huge or minuscule numbers, a methodology called scientific notation was born. 1.4.1

Very Large Numbers

To convert a large number like 324484738, for example, to scientific notation, we divide this number by 10, several times, until we get a nonzero number with just one digit in the integer part. Example: 324484738

10

=

32448473.8

32448473.8

10

=

3244847.4

3244847.38

10

=

324484.74

324484.738

10

=

32448.474

32448.4738

10

=

3244.8474

3244.84738

10

=

324.48474

324.484738

10

=

32.448474

32.4484738

10

=

3.2448474

1.4 Scientific Notation as a Tool

These numbers show that we have to divide the initial number 8 times by 10 to get a single digit alone in the integer part, like 3.24484738. For that reason, the given number in scientific notation is represented as 3.24484738 × 108. 1.4.2

Very Small Numbers

To convert a very small number to scientific notation, we must multiply the number by 10 several times, until we get a single-digit number in the integer part. The number of times the multiplication happens will be the exponent. In order to convert, for example, 0.00156 to scientific notation, it would be necessary to multiply it 3 times by 10 to obtain a single nonzero-digit number: 0.00156

×

10

=

0.0156

0.0156

×

10

=

0.156

0.156

×

10

=

1.56

For that reason, 0.00156 is represented in scientific notation as 1.56 × 10−3. Like we said before, negative exponents mean division. So, 1.56 × 10−3 can be also written as 1 56 1 56 = 103 1000 That is exactly 0.00156. 1.4.3

Operations with Powers of 10

Operations like multiplication, division, etc. can be done with numbers represented in powers of 10. This is how they are done. 1.4.3.1

Multiplication

To multiply two numbers, for example, 1.56 × 10−3 and 3.33 × 108, we must first multiply the numbers without the powers 1 56 × 3 33 = 5 19 and then sum the exponents −3 + 8 = 5 The result would be 5.19 × 105. If the final result is not a number with a single nonzero digit in the integer part, we must make sure it is.

3

4

1 Scientific Method

A final result of 24 × 1011 should be corrected to 2.4 × 1012. 1.4.3.2 Division

As explained before, any number raised to a negative exponent is equal to that number divided by the same power: 1 56 × 10 − 3 =

1 56 1 56 = 0 00156 = 103 1000

If we want to divide 3 × 109 by 8 × 102, we can represent that division as 3 × 109 8 × 102 The denominator can be converted to a nominator if we invert the exponent 3 × 109 = 3 × 109 × 8 × 10 −2 8 × 102 and we just converted a division into a multiplication. We can now multiply the numbers without the powers obtaining 8 × 3 = 24 and add the powers 9 + −2 = 7 and the final result will be 24 × 107, which should be converted to a single-digit number equal to 2.4 × 108. 1.4.3.3 Adding and Subtracting

To add or subtract numbers represented in powers of 10, we must first make the numbers have the same exponent. To add, for example, 2.4 × 108 and 3 × 109, we must first make both numbers have the same base. We chose 108. That will make the second number change to 30 × 108. We can now add both numbers: 2 4 + 30 = 32 4 The final result will be 32.4 × 108, corrected to 3.24 × 109. 1.4.3.4 Raising to a Number

To raise a number expressed in powers of 10 to a given exponent, we must raise the number without the power and then multiply the power by the exponent. The cube of 2.4 × 108 will be equal to the number alone raised to the cube 2 43 = 13 824

1.4 Scientific Notation as a Tool

and the base multiplied by the exponent 8 × 3 = 24 The final result will be 13.824 × 1024, corrected to 1.3824 × 1025. 1.4.3.5

Expressing Roots as Exponents

All roots in the form For example,

n

m

a m can be expressed as an exponent in the form a n .

1

3 = 32 3

1

7 = 73 5

85 = 82 1.4.3.6

Extracting the Root

To extract a root of a number expressed in powers of 10, we have to raise the number alone, without the power, to the correspondent exponent and multiply the power by that exponent. To calculate the square root of 2 4 × 108 we first have to write the number in the exponent form 2 4 × 108

1 2

We then extract the root of the number alone 2 4 = 1 549 and then multiply the base by the exponent 8×

1 =4 2

The result is 1.549 × 104. If the root is cubical, like 3

2 4 × 109

we first extract the cubical root of the number alone 3

2 4 = 1 338

and multiply the base by the equivalent exponent 9×

1 =3 3

5

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The result is 1.338 × 103. A problem arises if we need to extract a root of a number expressed in powers of 10 that is raised to another number, like 3

2 4 × 108

This root can be written as 2 4 × 108

1 3

As usual, we could start by extracting the root of the number alone and then multiplying the base by the exponent. But when we do that, we have an exponent that is 8 divided by 3, resulting in 108/3 or 102.666, which is not a very pretty result. To improve the result, we have to convert the exponent first to something that can be divided by 3. The closest candidate is 9. Hence, we convert 2.4 × 108 into 0.24 × 109, that is exactly the same number written differently. Now we can proceed. We first extract the cubical root of the number alone 3

0 24 = 0 6214465

and multiply the base by the exponent 9×

1 =3 3

The result is 0.6214465 × 103, which must be corrected to 6.214465 × 102. 1.4.4

Computers and Programming

Numbers written in scientific notation inside computer programs and other computer contexts use a specific syntax. A number like 1.38 × 1025, for example, is normally written as 1.38E25 or 1.38e25. Numbers with negative exponents have the sign before the number, for example, 1.38e − 25 or −3.43e − 12. In almost all the cases, computer languages use decimal numbers as fraction number separators. 1.4.5

Engineering Notation

In engineering, scientific notation numbers are expressed with exponents of 10 that are multiples of 3, so they match one of the multiples of the International System (see Appendix A).

1.4 Scientific Notation as a Tool

Following that rule, a number like 2 × 10−7 must be converted to 0.2 × 10−6 or 200 × 10−9. Normally it is always preferred to have numbers with first digits nonzero. For that reason, the later form will be generally preferred. In this case, the term 10−9 is equivalent to the prefix nano. So, the given number will be written as 200 nano or 200 n.

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2 Infinitesimal Calculus A Brief Introduction

2.1

Introduction

In this chapter we will do a brief introduction to infinitesimal calculus or differential and integral, known as simply, calculus. Wikipedia has as good definition about calculus: Calculus is the mathematical study of continuous change, in the same way that geometry is the study of shape and algebra is the study of generalizations of arithmetic operations. It has two major branches, differential calculus (concerning instantaneous rates of change and slopes of curves) and integral calculus (concerning accumulation of quantities and the areas under and between curves). These two branches are related to each other by the fundamental theorem of Calculus. Both branches make use of the fundamental notions of convergence of infinite sequences and infinite series to a well-defined limit. Generally, modern calculus is considered to have been developed, independently, in the 17th century by Isaac Newton and Gottfried Wilhelm Leibniz. Today, Calculus has widespread uses in science, engineering, and economics. Calculus is a part of modern mathematics education.

2.2

The Concept Behind Calculus

Suppose we have a car traveling in a straight line for 1 h, at a constant speed of 100 km/h, and later reducing the speed in half and traveling for another hour. What is the average speed of that car? Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

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The answer is 100 km h + 50 km h = 75 km h 2h Now let us see a more complex problem. The car travels 15 min at a nonconstant speed of 15 km/h, stops for 5 min, travels for 5 km at a nonconstant speed of 8 km/h, stops again, and then travels a distance of 2 km at 15 km/h. What is the average speed now? The answer is not evident because the method we have cannot deal with variable entities. Another method is required to perform this calculation. This is exactly the kind of calculation that is possible using infinitesimal calculus. 2.2.1

Limits

Limits are one tool of calculus that helps to calculate values that would be impossible by other methods. Consider the function f x =

4x− 4 4x− 4

Apparently, we can have the temptation to say that this function would be equivalent to 1, because both nominator and denominator are the same. The problem arises if we make x = 1, making the denominator equal to 0 and producing an indefinite result. If we plot the graph for this function, we get a horizontal line and an indefinite result when x = 1 (see Figure 2.1). Figure 2.1 Function plot.

2.2 The Concept Behind Calculus

We can say that the function is indefinite for x = 1, but how will the function behave for values very close to 1, like 0.9? If this is the case, we get f x =

4x −4 4x −4

f x =

4 0 9 −4 4 0 9 −4

f x =

−0 4 −0 4

f x =1 And what happens if x is even closer to 1, like 0.999999999999? The result will be the same, 1. The same will be true if for values of x are bigger but closer to 1, like 1.0000000000000001, then the result would continue to be 1. So, we conclude that the function is equal to 1, for any value closer to 1. It is like getting closer and closer to a dangerous point but never reaching it. This is the exact definition of limit. For the given function, we can say the same thing using the following mathematical notation: lim f x =

x

1

4x− 4 =1 4x− 4

the limit of f(x) when x approaches 1. 2.2.2

Derivatives

Derivatives are the second tool of calculus. Suppose we want to calculate the slope of the curve seen in Figure 2.2. Figure 2.2 A random curve.

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To calculate the slope, we can simply select two points at random, A and B, like shown in Figure 2.3, measure how much these points vary horizontally in X and vertically in Y, and find the slope by dividing the variation in Y by the variation in X. Figure 2.3 The slope of a curve.

Mathematically speaking, we should use the following formula for the slope: s=

ΔY ΔX

The slope of a line is easy, but imagine a function like f(x) = sin(x). How do we calculate the slope of this curve if the slope varies constantly in time? See Figure 2.4. Figure 2.4 Sinusoidal curve.

2.2 The Concept Behind Calculus

Obviously, it is clear that such curve has multiples slopes, depending on the point we select (see Figure 2.5). It is also clear that the curve has no slope per se. Only points belonging to that curve will have slopes.

Figure 2.5 Slope for a particular point on a curve.

Because the function varies in time, points on the curve will represent instantaneous values for a time in particular. To calculate the slope for a particular point on the curve, we can use the same method as before. We first define a second point on the curve, B, near the first one, as seen in Figure 2.6.

Figure 2.6 A second point on the curve.

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Now we can calculate the difference between both points, in terms of X and Y, as seen in Figure 2.7, and calculate the slope using the same formula as before. The problem with this kind of approach is that the result is just an approximation, because the formula we have calculates the slope of a line between A and B and the line is not the same as the curve, in terms of curvature.

Figure 2.7 The variation of X and Y.

We could improve the result by choosing a second point closer to the first one, as shown in Figure 2.8. By reducing the distance between the points, we make the line connecting these points closer, in terms of curvature, to the curve itself. The slope calculation now will give us a result closer to the real value. But we can improve that!

Figure 2.8 A line between point A and a second point closer.

If bringing the points together improves the result, how close can we get to have the correct result? The answer is: we can make the points infinitely close, and that is the exact definition of derivatives. Mathematically, the slope of point A can be defined as the difference in Y divided by the difference in X, which in other words is equal to

2.2 The Concept Behind Calculus

s=

f A + ΔY −f A ΔX

f(A) is the function value at point A. If we make ΔX = ΔY, we can use a generic Greek letter β to represent this variation. Thus, s=

f A + β −f A β

To get maximum precision, we have to make sure that the distance β is infinitely closer to 0, which will bring points A and B closer. We can, indeed, use the definition of limits to help us: s = lim β

0

f A + β −f A β

That is the slope of function f(x) as β gets closer to 0. 2.2.3

Integral

Integral is the third and last tool of calculus. To understand what Integral is, consider the following curve shown in Figure 2.9. Figure 2.9 A random curve represented on the X and Y axes.

Imagine this curve represents some kind of variation, like the temperatures of a city during a year, and we want to calculate the area below the curve, shaded area in Figure 2.10.

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Figure 2.10 Area below the curve.

To perform such calculation, we can use a technique similar to the one we used with derivatives. We can divide the area into rectangles, calculate the area of each rectangle, and sum them all, like shown in Figure 2.11. Figure 2.11 Rectangles being used to calculate the area under a curve.

Mathematically, the total area A equivalent to the sum of all small areas is equal to n

Ai

A= i=0

where Ai represents the areas of each rectangle. The area under the curve can be also written in terms of f(x): n

f x Δx

A= i=0

where f(x) is the function represented by the curve and Δx is the width of each individual rectangle.

2.2 The Concept Behind Calculus

Looking at Figure 2.11, it is clear that if we could create an infinite number of rectangles and sum all their areas, we would obtain the exact number representing the area below the curve. Mathematically, this area (A) can be expressed as a limit: n

f x Δx

A = lim n

∞

i=0

which is the exact definition of an integral, normally written in the following notation: A = f x dx

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3 Atom Quarks, Protons, and Electrons

3.1

Introduction

In this chapter, we will make a brief introduction about the physics behind electricity and electromagnetism, starting at the atomic level.

3.2

Atoms and Quarks

The word atom comes from ancient Greek word meaning “indivisible,” because it was believed at the time that the atom was the smallest part of matter and could not be divided. At the atomic level, atoms are composed of protons, neutrons, and electrons. Today we know that these elements are composed of more elementary elements called quarks, and known to exist in six variations called flavors: top, bottom, charm, strange, up, and down. As far as physicists know today, quarks are themselves elements that cannot be divided. Every time you try to divide them, a new quark is created. Protons for example, are composed of two up quarks and one down quark, and neutrons are composed of two down quarks and one up quark (Figure 3.1). Protons have a positive electric charge and neutrons have no electric charge.

Electric charge is normally represented, in physics, by the letter q or e.

In terms of reference charge, a proton is said to have an electric charge equal to 1. This charge, generally referred as e+, is equal to 1.602176634 × 10−19C.1

1 This is the new value revised by the SI in 2018. The old value was 1 6021766208 × 10 −19 C. Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

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Figure 3.1 A proton and its quarks.

3.3 Electrons Electrons are stable subatomic particles with a negative charge that can be found in all atoms and act like charge carriers. In terms of reference charge, an electron is said to have a charge equal to −1 or, in other words, a charge opposite to the charge of a proton. For that reason, its charge, generally referred as e−, is equal to −1.602176634 × 10−19C. In the early days, science pictured electrons spinning around atomic nucleus like planets spinning around the sun. Today we know that this image is false. Electrons oscillate frenetically around the atomic nucleus, and their position cannot be determined,2 like they were in all positions at the same time, a kind of “energetic” cloud. The idea of an electron orbiting an atomic nucleus like a planet around a sun is wrong because it makes us think about electrons as a solid and discrete element, like a particle. In real life, electrons behave both as particles and waves.3 Physicists knows that electrons occupy orbits around the nucleus at a certain distance, like they were shells. Each orbit is at a discrete level of energy and can only be occupied by electrons having that particular level of energy. If an electron gains energy, it may pass to a superior orbit. If it loses energy, it will decay to a lower orbit and release the extra energy by emitting a photon.4

2 Heisenberg’s uncertainty principle. 3 The famous wave–particle duality from quantum mechanics, demonstrated brilliantly in 1802 by Thomas Young, for the first time, by the double-slit experiment. 4 The photon is a type of elementary particle, the quantum of the electromagnetic field including electromagnetic radiation such as light, and the force carrier for the electromagnetic force. The photon has zero rest mass and always moves at the speed of light within a vacuum (source Wikipedia).

3.4 Strong Force and Weak Force

Figure 3.2 Atom representation.

A representation of an atom is shown in Figure 3.2. The spheres at the center represent the nucleus, nothing more than an agglomeration of protons and neutrons. The circles represent the energetic orbits. Electrons are shown in the diagram as “e” inside the orbits but in real life they can be at any point and in all points at the same time inside a particular orbit.

3.4

Strong Force and Weak Force

Atoms and their protons and neutrons are kept together by powerful attraction forces. The laws of physics say that elements with the same electric charge repel each other and that elements with opposite electric charge attract each other. Atomic nuclei may be formed by several protons and neutrons. These protons could never be kept together with other protons, at the required close proximity, to form these atomic nuclei, simply because they would repel each other. A bigger force must exist to overpower the electric repulsion force. This force is called the strong nuclear force or strong interaction. At a closer range, the strong force is hundreds of times as strong as electromagnetism, a million times as strong as the weak force, and a thousand times as strong as gravitation. The strong force keeps most ordinary matter together and binds neutrons and protons to create atomic nuclei. The weak force, weak nuclear force or weak interaction, takes place only at very small, subatomic distances, less than the diameter of a proton. It is one of the four known fundamental interactions of nature, alongside the strong interaction, electromagnetism, and gravitation. This weak force is responsible for radioactive decay and thus plays an essential role in nuclear fission.

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3.5 Conductors and Electricity As mentioned before, atoms are composed of protons, holding a positive electric charge and neutrons that have no electric charge. Electrons are found frenetically oscillating in a cloud of energy around this agglomerate of protons and neutrons. We know that electrons are charge carriers and that as they move inside a material, they transport electric charges around. This movement of charges is called electric current. In order for an electron to be able to move around a material, it is necessary, for the material itself, to have a certain kind of atomic structure. Hydrogen (H), for example, is the simplest atomic element and for that reason the most abundant element in the universe. Hydrogen’s nucleus is formed of a single proton and one electron is found orbiting around (see Figure 3.3). Figure 3.3 Hydrogen atom.

Helium (He), on the other hand, is composed of two protons, two neutrons, and two electrons, like shown in Figure 3.4.

Figure 3.4 Helium atom.

3.6 The Shells

Hydrogen and helium are not good conductors of electricity. They have very few electrons, and these electrons are closer to the nucleus, subjected to the strong force, and, for that reason, are very difficult to move out of the atomic structure. Copper (Cu), for example, an excellent conductor of electricity, is much more complex than hydrogen and helium. Its nucleus contains 29 protons, 35 neutrons, and 29 electrons. The electrons are distributed in several energetic shells, as represented in Figure 3.5. Figure 3.5 Copper atom.

Copper contains 29 electrons disposed in four shells. From inside out, these shells have 2, 8, 18, and 1 electrons, respectively.

3.6

The Shells

Atoms that have a lot of electrons have these electrons distributed through several orbits, or shells, around the nucleus, like shown in Figure 3.5. An electron needs a certain level of energy to occupy a particular orbit. Orbits that require more energy are farthest from the nucleus. Each orbit, or shell, allows a certain limited number of electrons. The first six shells, for example, known by the letters K, L, M, N, O, and P, can have 2, 2, 18, 32, 50, and 72 electrons, respectively. The outermost shell of an atom is known as the valence shell. The number of electrons allowed by the atomic shells follows a mathematical pattern equal to 2n2, where n is the shell number, starting with 0.

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Electrons are not allowed to exist between shells. If an electron is on a particular shell and absorbs the required energy level to jump to an outer shell, it does by disappearing from the current shell and appearing on the outer shell. If it is the other way around, that is, an electron is on an outer shell and loses energy to be on a vacant position in an inner shell, it disappears from the current shell and appears on the inner shell. In the process, it releases the extra energy by emitting a photon. Electrons that are on the outermost shells are less subject to the strong force, exerted by the nucleus. Hence, these electrons can be more freely moved, if energy is applied. Imagine a block of copper, containing millions and millions of atoms of this element. When these atoms are packed together, a magic happens and something known as a metallic bonding is formed, leaving a huge number of electrons detached from the nucleus. These electrons are now completely free to move if energy is applied.

3.7 Electric Potential Like explained before, protons have positive electric charge and electrons have negative electric charge. A large number of electrons on one side of a given material compared with another represent a difference in electric potential between the two points. This difference is electric potential energy stored in the form of electric charge. A difference in electric potential between two points will create an electric field, and this field will force electrons to move into a specific direction. The bigger the difference in potential, the bigger is the electric field, and, consequently, the bigger the number of electrons moving, the bigger is the current. A large difference in potential will also cause electrons to move more aggressively through the field. The difference in electric potential is a kind of potential energy and as an energy it can create work. The electric potential energy can be defined as the work, in Joules, to move electrons on an electric field. The difference in electric potential, also known as electromotive force, is measured in Volts5 and is represented by the uppercase letter V in the SI.6

5 In honor of Alessandro Giuseppe Antonio Anastasio Volta, simply known as Alessandro Volta (1745–1827), an Italian physicist, a chemist, and a pioneer of electricity and power who is also credited as the inventor of the electric battery. 6 French’s Système Internationale d’Unités (International System of Unities), abbreviated as SI in all languages.

3.9 Electric Resistance

3.8

Current

Like mentioned previously, an electric field applied to a material will create a kind of pressure that will force the electric field to be transmitted from electron to electron at a speed about 63% of the light speed. The electrons themselves move very slowly, from the lower potential point to the higher potential point, at a speed, called “drift velocity” that is something like 1.5 cm/min. The movement is irregular, like in a random zigzag motion, as they collide with atoms in the conductor. The effect of having the electric field transmitted at a speed higher than the electron speed can be illustrated, roughly, to what happens on the Newton cradle toy (see Figure 3.6). Figure 3.6 Newton cradle.

All balls are at initially at rest until we apply a speed to the first one in the left and make it collide with the other ones. As soon as the collision happens, the last ball in the right jumps. In other words, the balls practically do not move, but the force is transmitted from one to the other through the entire chain. Current is measured in Ampere7 and is represented by the uppercase letter A in the SI.

3.9

Electric Resistance

Free electrons inside a material can move forced by an electric field, but not all electrons are completely free to move, due to restrictions imposed by the material’s structure. These restrictions prevent the free movement of electrons and resist their passage, generating friction and heat. This restriction is called electric resistance, measured in Ohm,8 and is represented by the Greek letter omega (Ω) in the SI. 7 In honor of André-Marie Ampère (1775–1836), French mathematician and physicist, credited as the father of electrodynamics. 8 In honor of Georg Simon Ohm (1787–1854), German physicist and mathematician.

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4 Voltage and Current Direct and Alternating Current and Voltage

4.1

Introduction

In the last chapter we have described the characteristics behind difference in electric potential, electric fields, and current. In this chapter, we will describe direct current (DC) and alternating current (AC) and voltage.

4.2

Terminology

The term “direct current” is used indiscriminately to refer to direct voltage and to direct current. In this book we will differentiate the terms to prevent confusion. The same thing is valid for “alternating current,” which in the real world is used to refer to both alternating voltage and alternating current. Other terms also used are VDC to refer to “DC voltage” or direct (continuous) voltage and VAC to refer to “AC voltage” or alternating voltage.

4.3

Batteries

Around 1800, Alessandro Volta reported the results of an experiment where he was able to create electricity by piling discs of zinc and copper separated by felt spacers soaked in salt water. With this invention, Volta proved that electricity could be generated chemically and debunked the prevalent theory that electricity was generated solely by living beings. A new invention was born: the battery.

A device that is able to generate electricity is generally referred as power supply.

Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

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4 Voltage and Current

4.3.1

Battery Life

Batteries are electrochemical devices created by embedding two different metallic plates into a third one. This third one, the electrolyte, is responsible, broadly speaking, for creating a chemical reaction in such a way that one of the elements ends with excess of electrons (negative charges) and the other one with an excess of positive charges – in other words, to create an imbalance of charges between the plates. To summarize how chemical energy is converted into electrical energy, let us take an example of a typical car battery: one plate of pure lead and another one made of lead dioxide both immersed in a solution of sulfuric acid diluted with water. Chemical reaction causes the lead to become negatively charged and the lead dioxide to become positively charged. The element with excess of electrons will be the negative pole (cathode) and the other element will be the positive pole (anode). This difference in the number of electrons between the poles is called “difference in electric potential” or popularly, “voltage.” Any element, like a light bulb, for example, connected between the poles, will provide a path for the excess of electrons of the lead element to migrate to the lead dioxide, which lacks electrons. As the reaction happens, most of the sulfuric acid in the electrolyte is consumed and used to convert both plates into lead sulfate, and the electrolyte turns into, mostly, regular water. At some point, all three elements are electrically balanced, current stops flowing, and the battery dies. In electricity’s early days, voltage was known as electromotive force (EMF), because voltage is considered to be a kind of “pressure” that pushes the electric field through a conducting loop. It is known that electrons flow from the negative to the positive pole, at a certain slow speed; when compared with the speed, the electric field travels in the same direction. Current is said to flow in the opposite direction, from positive to negative, because its direction was established before the electron discovery.

The process of charging the battery will apply energy to the plates and electrolyte to reverse the reaction, making the plates to be lead dioxide and lead and the electrolyte sulfuric acid and water, again.

All power supplies have a limited voltage and current capacity. Current has a direct relation to the number of electrons that a power supply is able to move over time.

4.3 Batteries

4.3.2

Batteries in Series

Consider two identical batteries connected in series, like shown in Figure 4.1. The positive pole of one battery is connected to the negative of the other and two wires are connected to points A and B. Figure 4.1 Batteries in series.

If each battery has a voltage of 1.5 V, two batteries connected in series will provide a combined voltage of 3 V. This will be the value measured across A and B. Batteries can only provide a limited amount of current. If each battery on the given example can provide 0.2 A of current, connecting any number of batteries in series will not change that. Conclusions:

••

Two power supplies in series will have their voltages added. Two power supplies in series will not have their currents added.

4.3.3

Batteries in Parallel

Figure 4.2 shows two identical batteries connected in parallel. The positive poles and negative poles are connected together. Figure 4.2 Batteries in parallel.

Supposing each battery can provide 1.5 V and 0.2 A,1 the voltage across AB will be 1.5 V, and together the batteries will be able to provide 0.4 A of current, which is the sum of their individual currents. 1 This can be written as 1.5 V @ 0.2 A.

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4 Voltage and Current

••

Conclusions: Two power supplies in parallel have their currents added. Two power supplies in parallel will not have their voltages added.

4.4 Danger Will Robison, Danger! When connecting batteries in series or parallel, a few rules must be followed. 4.4.1

Never Invert Polarities

Inverting a battery in a block of batteries in series, like shown in Figure 4.3, or in parallel may represent a danger situation. Figure 4.3 Incorrect connection.

The other batteries will force current circulation across the inverted one, generating excessive heat that may lead to fire hazard and explosions. 4.4.2

Never Use Different Batteries

A battery is only capable of providing a limited amount of current and voltage. If you connect batteries with different current capabilities in series, for example, 2, 4, and 6 A, the largest battery will force 6 A of current through batteries that cannot handle that amount of current. Excessive heat may be generated, again creating a danger situation. The same is true for batteries with different voltages in parallel. 4.4.3

Short-Circuiting Batteries

Never short-circuit a battery like shown in Figure 4.4. Current will flow between them, at maximum level, making both batteries explode and catch fire.

4.6 Relative Voltages

Figure 4.4 Batteries in short circuit.

4.5

Direct Current

Direct current or DC is a term used to refer to the kind of voltage or difference in potential that keeps its value constant over time and does not vary in amplitude or frequency. Figure 4.5 shows the symbol used to refer to a source of DC. In that case, the symbol is used to refer to batteries. Figure 4.5 Battery symbol.

Batteries are sources of DC, meaning that once created they will provide a constant voltage and current during its lifetime. 4.5.1

•• •• •

DC Characteristics

Current and voltage are constant over time. They are generated by chemistry and are very difficult to deploy in large scale. They never change polarity over time. Current only flows in one direction during all time. A DC voltage is not able to cross large distances. They lose power, considerably, if transmitted over long distances.

4.6

Relative Voltages

Consider two different electric potentials, like two batteries in series. How relative is the potential of one battery in respect with the other?

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If each battery provides 1.5 V and two batteries in series provide 3 V, what are the references to which we are measuring these numbers? What potential is positive and what potential is negative? Is a potential really positive or negative? In relation to what? 4.6.1

Mountains

To understand why, consider two mountains like shown in Figure 4.6 and a path that one can climb from A to C, including a point B in the middle.

Figure 4.6 Two mountains.

A person standing on point A can climb to C in two phases: from A to B, resting at B a little bit, and then from B to C. In terms of physics, the climber is going against gravity and, by doing so, accumulates gravitational potential energy. Gravitation potential energy is calculated by the following formula. POTENTIAL ENERGY

Egp = mgh Egp is the gravitational potential energy, in Joules.2 m is the object’s mass, in kilograms. g is the gravity acceleration on Earth, equal to 9.834 m/s2. h is the vertical distance traveled by the object, in meters.

If point A is at sea level, B is at 100 m, C is at 200 m, and a person weights 70 kg, we can use the given formula and calculate that the person will gain a potential energy of 68.6 kJ by climbing from A to B and the same amount, again, by climbing from B to C. If we choose point A as zero reference, we conclude that the person will have a potential energy of zero at point A, 68.6 kJ at B, and 137.2 kJ at C. 2 In honor of James Prescott Joule (1818–1889), English physicist. Joule is equivalent to kg.m2/s2 and represented by the uppercase letter J in SI.

4.7 Ground

What happens if instead of point A we select point B as zero reference? If we do, we conclude that point C is 68.6 kJ over B and A is 68.6 kJ under B. In other words, in relation to B, the energy at C is 68.6 kJ and the energy at A is −68.6 kJ. The negative sign is used to indicate that the potential level of A is below the reference point. Climbing down from B to A means losing potential energy, because we walk from a point with a higher potential energy to another with a lower potential energy.

Now consider two batteries in series, like shown before, but, this time, we take them apart to include a third point B, like shown in Figure 4.7. Figure 4.7 Batteries in series with an intermediary point.

Figure 4.7 shows three points: A, B, and C. No need to remember the mountain analogy. If each battery can provide a voltage of 1.5 V and we use point A as zero reference, we see that point B will be 1.5 V and point C will be 3 V above A. On the other hand, if we use point B as a zero reference, C will be at 1.5 V above B and A will be −1.5 V below B. Again, the negative sign is used to show that A is below B. This example shows the concept of negative electric potentials that are nothing more than a way to tell if a voltage is above or below a zero reference. This zero-reference point is called ground point, or simply ground, or in some cases, earth point.

4.7

Ground

In an electric circuit, the zero reference is called ground. The most used abbreviation in schematics is GND and its symbols are shown in Figure 4.8. Figure 4.8 Ground symbols.

33

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4 Voltage and Current

4.8 Alternating Current AC is a term used to refer to an alternating voltage or to an alternating current source. An alternating voltage changes its amplitude values constantly over time in a repeating pattern. Like the alternating voltage, an alternating current also changes its amplitude values over time as well but additionally inverts polarity in a repeating pattern.

When current reverses polarity, it means a change in direction.

Figure 4.9 shows the symbol used for AC voltage sources in schematics.

Figure 4.9 Alternating current voltage source symbol.

4.8.1

•• •• •

AC Characteristics

Can be easily created in large or small scale with electric generators. The levels of voltage and current vary cyclically with time. Voltage polarity may change with time. Current direction changes with time. Can travel large distances over wires with minimum losses.

4.8.2

AC Cycles

Figure 4.10 shows a graph representing an alternating voltage. In Figure 4.11, we can see that alternating waves have a positive cycle and a negative cycle and vary over time in intensity. The wave starts at 0 V, increases until it reaches V+, and then decreases to a negative level V−, repeating the whole pattern after that, ad infinitum. Because the alternating voltage follows a cyclic pattern going up and down constantly, over time, it oscillates at a specific constant frequency. This

4.8 Alternating Current

Figure 4.10 Alternating voltage.

Figure 4.11 Alternating current cycles.

frequency is measured in Hertz,3 meaning cycles per second,4 and is represented in the SI by the letters Hz. 4.8.3

Period and Frequency

Figure 4.12 shows that the alternating wave starts at 0 and goes up and down and after a period equal to T starts repeating it. In the given example, T is equal to 4 s and is known as the wave’s period of oscillation. The frequency and the period of a wave are related by the following formula. 3 In honor of Heinrich Rudolf Hertz, the first to prove, conclusively, the existence of electromagnetic waves. 4 A cycle is the period of time a wave needs to complete itself. Old schematics and books used the term cycles per second (cps) as a way to refer to wave frequency. It is not rare to find old texts referring to frequency as kcps (kilo cycles per second), instead of Hz.

35

36

4 Voltage and Current

Figure 4.12 Period of a sinusoidal wave.

WAVE FREQUENCY

f=

1 T

f is the wave frequency, in Hertz. T is the wave period, in seconds.

4.8.4

Peak-to-Peak Voltage

Consider the wave shown in Figure 4.13, where voltages vary from 20 to −20 V. These values are measured against the 0 and are called peaks.

Figure 4.13 Alternating current peaks.

4.8 Alternating Current

Remembering the mountain analogy, we can now use the negative peak as zero reference. If so, we can say that the positive peak is 40 V above that. In other words, 40 V is the difference between both peaks, also known as peak-to-peak voltage.

Peak-to-peak voltage is generally represented by the abbreviation VPP.

4.8.5

DC Offset

We have described DC and AC as two separate entities, but we never said that they cannot exist together. Observe the alternating voltage shown in Figure 4.14. The wave reference rests, vertically, at 0 V. The wave varies from 20 to −20 V. Figure 4.14 40 Vpp Alternating voltage.

Now suppose we add 10 V continuous to that wave. The result is seen in Figure 4.15. Figure 4.15 Alternating current plus 10 V DC.

37

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4 Voltage and Current

By observing Figure 4.15, it is clear that the wave shifted up by 10 V. Now, its positive peak is at 30 V and its negative peak is at −10 V. The wave now rests, vertically, at 10 V and not at 0 V as before. We say, in that case, that an offset of 10 V was applied to the alternating voltage. If the offset is negative, −10 V, for example, the wave of Figure 4.14 will shift down and be like the one shown in Figure 4.16. The wave now rests at −10 V and variates up and down to 10 V and −30 V, respectively. In other words, we pushed the wave down by 10 V. Figure 4.16 Alternating current minus 10 V DC.

Exercises 1

Consider the circuit shown in Figure 4.17, composed of several power supplies. Find the voltage between points A and B. Figure 4.17 Batteries in series (Exercise 1).

Solutions

2

Consider the same circuit shown on Figure 4.17. The three batteries of 3 V, 12 V, and 6 V are capable of providing 2 A, 4 A, and 3 A, respectively. Find the maximum current this circuit can provide.

3

Circuit shown on Figure 4.18 is composed of three power supplies. Find the voltage between points A and B. Figure 4.18 Batteries in series (Exercise 3).

4

Consider a 12 V alternating voltage power supply in series with an 8 V direct voltage power supply. What are the resulting peak voltages of this combination?

5

Find the period of oscillation of a 60 Hz AC power supply.

Solutions 1

If we follow the circuit from point B to A, we cross the 3 V power supply from the negative to the positive pole. It means that we are going from a point of low electric potential to a point of high electric potential and consequently we went up 3 V. If we continue to follow the circuit, we cross the 12 V battery, again, from the negative to the positive pole, increasing the potential an additional 12 V. The same will happen to the third battery, raising the potential an additional 6 V. So, the voltage across AB is the sum of these three potential increases.

FINAL RESULT

VAB = 3 + 12 + 6 = 21 V

39

40

4 Voltage and Current

2

Batteries with different current capabilities should never be connected in series and may lead to fire hazard or explosion.

3

If we follow the circuit from B to A, we will cross the first voltage source (3 V) from its negative to its positive pole, a 3 V increase in potential. As we continue to follow the circuit, we cross the second battery (12 V), from the positive to the negative pole, meaning a potential decrease of 12 V. As we cross the third battery (6 V), we see an increase of 6 V. So, the voltage across AB is equal to

FINAL RESULT

VAB = 3 + 6 − 12 = − 3 V

The negative sign means that B’s potential is lower than A’s. 4

If a source of alternating voltage has an amplitude of 12 V, it means it can reach a positive peak of 12 V and a negative peak of −12 V. An 8 V direct voltage source added to this alternating source will shift the wave 8 V up. So, the wave will now reach a positive peak of 20 V(12 + 8) and a negative peak of −4 V(−12 + 8).

5

The period of a wave is given by T= T=

1 f

1 60

FINAL RESULT

T = 0 0167 s

41

5 Resistors The Most Fundamental Component

5.1

Introduction

This chapter is about resistors, the most basic electric component.

5.2

Resistor

Resistors are the most fundamental and commonly used of all the electronic components. Having two terminals, this component opposes current flow. This opposition is called electric resistance, measured in Ohms, and is represented by the Greek letter omega (Ω) in the SI.

5.3

Electric Resistance

All materials offer, to a certain degree, resistance to the flow of electrons. Metals, for example, offer very little electric resistance, but the resistance is there and it is not 0. Factors like surface area, thickness, and chemical properties can alter electric resistance, and by manipulating these characteristics correctly, several types of resistors can be created.

5.4

Symbols

There are two different symbols to represent a resistor in schematics, one in the American standard and the other in the international standard. Figure 5.1 shows both symbols.

Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

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5 Resistors

Figure 5.1 Resistor symbol (US and international standard).

Figure 5.2 Real standard resistors.

Figure 5.2 shows a picture of real resistors where color stripes describing their resistance values and other characteristics can be seen around their bodies.

5.5 Types of Resistor There are several types of resistors produced using a variety of technologies: wire, carbon, metallic oxide film, etc. Every type of resistor is useful for a specific application. There are resistors suitable for high stability, high voltage, high current, and high precision, among other uses. Resistors have different degrees of precision called “tolerance.” The more precise or more suitable for a specific kind of application, the higher the cost. General-purpose cheaper resistors may have a tolerance of 20%, for example, meaning that a resistor of 1000 Ω, for instance, is only guaranteed to have a real resistance between 800 and 1200 Ω.

5.6 Power Resistors are produced to handle a certain amount of current and, consequently, a certain amount of heat and power in Watts.1 Typical values for power are 1/8 W, 1/4 W, 1/2 W, 1 W, etc.

5.7 Color Code Standard regular resistors have their characteristics encoded in four color stripes around their bodies. 1 We will talk about power in a future chapter.

5.7 Color Code

The first two stripes define the first two digits of their resistance, the third represents the multiplier, and the fourth represents the tolerance. See Table 5.1.

Table 5.1 Resistor color code. Color

First stripe

Black

Second stripe

Multiplier

0

×1

Tolerance (%)

Brown

1

1

× 10

±1

Red

2

2

× 100

±2

Orange

3

3

× 1000

Yellow

4

4

× 10k

Green

5

5

× 100k

± 0.5

Blue

6

6

× 1M

± 0.25

Violet

7

7

× 10M

± 0.1

Gray

8

8

× 100M

± 0.05

White

9

9

× 1G

Gold

× 0.1

±5

Silver

× 0.01

± 10

A resistor with the stripes Brown

1

Black

0

Yellow

×10k

Gold

±5%

will be equal to 10 × 10000 = 100 kΩ with a tolerance of ±5%. Some types of resistors may have a fifth stripe. If so, the first three stripes represent the value, the fourth is the multiplier, and the last is the tolerance.

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5 Resistors

A resistor of this kind, with the stripes shown next, will have the following characteristics: Orange

3

Orange

3

Black

0

Brown

×10

Violet

±0.1%

The resistor will be 330 × 10 = 3.3 kΩ with a tolerance of ±0.1%. Resistors in the range of kΩ and MΩ are also identified by using uppercase letters K and M, like 1K2 and 2M7 to identify 1.2 kΩ and 2.7 MΩ, respectively. On the other end, the uppercase letter R is used to represent very small values of resistance. A resistor of 1.2 Ω may be referred to as 1R2. We have compiled a table shown in Appendix B, which makes easy to find resistance values for color-coded resistors. Appendix G shows several tables containing the values of resistances available in the market and their tolerances.

5.8 Potentiometer Regular resistors have two terminals and a fixed value of resistance. Many situations in real life require resistors that can vary their resistance. For this reason, a new kind of resistor was born: the potentiometer. The potentiometer is basically a three-terminal device with a central axis that can be rotated or moved to make between its terminals. The volume rotary device used in amplifiers is one example. Symbols for the potentiometer in the American and international standards can be seen in Figure 5.3.

5.9 Trimpots Trimpots or trimmer potentiometers are miniature potentiometers used for adjustment, tuning, and calibration in circuits. The resistance of trimpots and potentiometers can vary linearly or logarithmically. Figure 5.3 Potentiometer symbols (US and international standard).

5.12 Resistors in Series

5.10

Practical Usage

Incandescent light bulbs are examples of resistors with a very small value of resistance. When used in circuits, resistors are used to control the flow of current. This property can be used to protect sensitive components from excessive current. Other uses involve dividing voltages and creating timers in conjunction with capacitors.

5.11

Electric Characteristics

Resistors are “passive components,” because they have no active role in a circuit. They simply resist current flow. As they operate, they reduce current, and by doing so, they create an electric potential difference between their terminals. A resistor may have 10 V on one terminal and just 8 V on the other. For that reason, it is said that resistors create voltage drops across their terminals. 5.11.1

Important Facts

Obviously, voltage drops will just happen if current flows across a resistor. If current is 0, both terminals will have the same voltage. Even without any current, resistors have what is called thermal noise, related to the internal chaotic movement of electrons inside the material’s structure. When current flows across a resistor, it creates friction against its internal structure, generating heat and increasing the thermal noise. This noise leaks to the circuit and may interfere with other components, distort signals, and create other undesired problems. Different types of resistors were created to minimize problems like this.

5.12

Resistors in Series

Resistors connected in series, like shown in Figure 5.4, will have a total resistance equal to the sum of each individual resistance, or

Figure 5.4 Resistors in series.

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5 Resistors

R = R1 + R2 For an infinite number of resistors, the following formula applies. EQUIVALENT RESISTANCE: SERIES R = R 1 + R2 +

+ R∞

R is the equivalent resistance, in Ohms. R1 , R2 ,… are the values of the individual resistors, in Ohms.

5.13

Resistors in Parallel

Resistors connected in parallel, like shown in Figure 5.5, will have a total resistance that can be found by the following formula: 1 1 1 + = R R1 R2

Figure 5.5 Resistors in parallel.

If we have only two resistors, we can use this other friendly formula: R=

R1 × R2 R1 + R2

For an infinite number of resistors, we get the following formula. EQUIVALENT RESISTANCE: PARALLEL 1 1 1 + + = R R1 R2

+

1 R∞

R is the equivalent resistance, in Ohms. R1 , R2 ,… are the values of the individual resistors, in Ohms.

5.14

DC and AC Analysis

Differently from other components we will still examine in this book, resistors are the only components that behave the same way when connected to direct current or to alternating current.

5.15 Input and Output Synchronism

Figure 5.6 A simple resistive circuit.

Suppose we connect points A and B of the circuit shown in Figure 5.6, to a battery. Current flows across the several components in the circuit and a certain voltage would be measured across points C and D. This voltage would be less than the one applied between A and B. If the voltage across AB varies, voltage across CD will vary instantly. If we replace the battery with a source of alternating voltage, the result will be the same in essence: we will measure a voltage between C and D that is less than the one applied to the input. Again, if the voltage across AB varies, voltage across CD will vary instantly.

5.15

Input and Output Synchronism

Imagine the circuit in Figure 5.6, initially at rest. No input or output voltage. At time equal to 0, we apply an alternating voltage at the circuit’s input and start monitoring what happens at the output. We will call t0 this instant in time. At the time the voltage is applied across the input, a smaller voltage appears across the output. As the input voltage varies, the output voltage also varies, at the same time, without any delay. We say that both input and output are in synchronism or in phase or have a phase equal to 0 . In Figure 5.7 we show an input signal (larger wave) and the respective output signal (smaller wave) in synchronism. If we want to make an analogy to the real world, we could say that resistors have zero inertia and react instantly to any variations Figure 5.7 Input and output signals in phase. in voltage or current.

47

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5 Resistors

Exercises 1

Consider three resistors in series: 220 Ω, 180 Ω, and 1 kΩ. What is the equivalent resistance?

2

Consider three resistors in parallel: 1.2 kΩ, 100 kΩ, and 1 Ω. What is the equivalent resistance?

3

Figure 5.8 shows a purely resistive circuit. What is the resistance measured between points A and B? Figure 5.8 Resistive circuit (exercise).

4

A resistive circuit is connected to an alternated voltage source. What is the phase angle between output and input?

Solutions 1

The equivalent resistance of resistors in series is the sum of them all. Hence, REQ = R1 + R2 + R3 REQ = 220 + 180 + 1000

FINAL RESULT REQ = 1400 Ω

2

The equivalent resistance of resistors in parallel is found by the formula 1 1 1 1 = + + REQ R1 R2 R3 1 1 1 1 = + + REQ 1200 100000 1

Solutions

FINAL RESULT REQ = 0 9992 Ω

3

Resistors R1 and R2 are in series. Their equivalent resistance (REQ1) will be the sum of their individual resistances: REQ1 = R1 + R2 REQ1 = 1800000 + 2 2

REQ1 = 1800002 2 Ω The remaining resistors, R3, R4, and R5, are in parallel. Their equivalent resistance follows the formula 1 1 1 1 = + + REQ2 R3 R4 R5 1 1 1 1 = + + REQ2 10000 22000 47000

REQ2 = 5997 6769 Ω REQ1 and REQ2 are in series. We must sum them to obtain the total resistance between points A and B: RAB = REQ1 + REQ2 RAB = 1800002 2 + 5997 6768

FINAL RESULT RAB = 1805999 88 Ω

4

Pure resistors do not affect the phase of a signal.

FINAL RESULT Phase = 0o

49

51

6 Ohm’s Laws Circuit Analysis

6.1

Introduction

In this chapter we will examine Ohm’s law, postulated by German physicist Georg Simon Ohm in 1827, that relates resistance, current, and voltage in conductors.

6.2

Basic Rules of Electricity

A few basic rules of electricity are always true, independent of which components we connect in a circuit:

• • •

Components in parallel will always have the same voltage and divide the current. Components in series will always have the same current and divide the voltage. Components with the same voltage in both terminals do not have any current flowing across.

Consider the circuit shown in Figure 6.1. Figure 6.1 shows three resistors in parallel. These resistors have the same voltage across. Current, on the other hand, will divide between them. Figure 6.1 Resistors in parallel with a battery.

Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

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6 Ohm’s Laws

If all resistors have the same resistance, each one will receive one-third of the total current. On the other hand, if they have different resistances, each one will have a current inversely proportional to their resistance. Anyway, the total current will be always equal to the sum of the currents of all parallel branches. In Figure 6.2 we show another circuit. This time, three resistors are in series with a battery. Figure 6.2 Resistors in series with a battery.

In the circuit shown in Figure 6.2, all resistors have the same current across, because they are in series. On the other hand, they will divide the battery’s voltage among them. If all resistors have the same resistance, each one will have one-third of the voltage provided by the battery. On the other hand, if they have different resistances, their voltages will depend on a relation between their resistances and the sum of all three resistances. Anyway, the sum of their voltages will always be equal to the voltage provided by the battery.

6.3 First Ohm’s Law The first Ohm’s law postulates that the current across two points of a conductor is directly proportional to the voltage across these same two points, according to the following relation. FIRST OHM’S LAW V = RI V is the voltage, in Volts. R is the resistance, in Ohms. I is the current, in Amperes.

6.5 Examples

6.4

Second Ohm’s Law

The second Ohm’s law states that the resistance of a homogeneous conductor of constant cross section is directly proportional to its length and is inversely proportional to the area of its cross section. SECOND OHM’S LAW R=

ρL A

V is the voltage, in Volts. ρ is the conductor resistivity, in Ohm meters (depends on the material and its temperature). L is the conductor length, in meters. A is the conductor cross section, in squared meters.

6.5

Examples

6.5.1

Example 1

Figure 6.3 shows a circuit where a 1000 Ω resistor is connected to a 12 V battery. We want to find the current flowing in the circuit. 6.5.1.1

Solution

The firs Ohm’s law postulates that V = RI

Figure 6.3 One resistor in series with a battery.

Hence, current will be I=

V 12 = R 1000

FINAL RESULT I = 0 012 A = 12 mA

6.5.2

Example 2

Figure 6.4 shows a circuit where two resistors are in parallel with a battery. The battery provides 12 V and the total current drained by the circuit is 3 A. We want to find the current flowing across each resistor.

53

54

6 Ohm’s Laws

Figure 6.4 Two resistors in parallel with a battery.

6.5.2.1 Solution

We know that the total current I is equal to the sum of the branch currents, I1 and I2, that are the currents flowing across R1 and R2, respectively: I = I1 + I2 R2 is 500 Ω, exactly half the resistance of R1, which is 1000 Ω. Thus, R2 will receive twice the current of R1: I2 = 2I1 We know that I = I1 + I2 Upon substituting the first condition, I = I1 + 2I1 I = 3I1 The total current I is equal to 3 A; thus 3I1 = 3 Hence FINAL RESULT I1 =

3 = 1A 3

To find I2, we know that I2 = 2I1 Hence,

6.5 Examples

FINAL RESULT I2 = 2 A

6.5.2.2

Another Method to Obtain the Same Result

R1 and R2 are in parallel; thus they have the same voltage. We know that the voltage across a resistor is V = RI Thus, R1 I1 = R2 I2

This formula tells us that the voltage across R1 (R1 I1 ) is equal to the voltage across R2 (R2 I2 ).

Hence, I1 =

R 2 I2 R1

Upon substituting the values, we get I1 =

500 × I2 1000

I1 = 0 5 I2 This is the same conclusion we have found before, that is, I2 is twice I1. If I2 is twice I1, we can divide the total current into three equal parts and designate two parts for I2 and one for I1. FINAL RESULT I1 = 1 A I2 = 2 A

6.5.3

Example 3

Consider the circuit shown in Figure 6.5, containing two resistors in series with a battery.

Figure 6.5 Two resistors in series with a battery.

55

56

6 Ohm’s Laws

The battery provides 12 V, and the resistors, R1 and R2, have resistances of 1000 Ω and 4700 Ω, respectively. We want to find the total current flowing in the circuit. 6.5.3.1 Solution

Like we have learned before, the equivalent resistance of resistors in series is the sum of their individual resistances. Thus, the total resistance is 5700 Ω. By Ohm’s law, V = RI Upon substituting the values, the current flowing in the circuit is 12 = 5700 I I=

12 5700

FINAL RESULT I = 0 0021052 = 2 1052 mA

6.5.4

Example 4

Using the same circuit shown in Figure 6.5, find the voltage drop across R1. 6.5.4.1 Solution

A current flowing in a resistor produces a voltage drop, which can be calculated by using V = RI We know that the total current, flowing across both R1 and R2, is 2.1 mA. Therefore, the voltage drop across R1 will be its resistance multiplied by the current flowing across itself: VR1 = R1 I VR1 = 1000 × 0 0021052

FINAL RESULT VR1 = 2 1052 V

6.5 Examples

6.5.4.2

Another Method to Obtain the Same Result

We know that R2’s resistance is 4700 Ω and that current flowing across that resistor is 2.1 mA. By Ohm’s law, V = RI Thus, voltage drop across R2 is VR2 = R2 I VR2 = 4700 × 0 0021052

VR2 = 9 8947 V R1 and R2 are in series. The voltage across both resistors is 12 V. So, the sum of the voltage drops across each resistor must be also equal to 12 V: VR1 + VR2 = 12 V Hence, VR1 = 12−VR2 VR1 = 12−9 8947

And we obtain the same value as before. FINAL RESULT VR1 = 2 1052 V

6.5.5

Example 5

Suppose we have 100 m of wire with a cross section equal to 5 mm2. We apply 200 V to this wire and measure a current of 10 A. What is this conductor’s resistivity? 6.5.5.1

Solution

We use the first Ohm’s law to find the wire’s resistance: V = RI Thus, V 200 = = 20 Ω I 10 By the second Ohm’s law, R=

57

58

6 Ohm’s Laws

R=

ρL A

ρ=

RA L

or,

Substituting the values, we get R = 20 Ω A = 5 mm2 = 5 × 10 −6 m2 L = 100 m ρ=

20 × 5 × 10 −6 = 1 × 10 − 6 Ω m 100

FINAL RESULT ρ = 1 μΩ m

Exercises 1

Consider the circuit shown in Figure 6.6, composed of three resistors and a battery.

Figure 6.6 Resistive circuit and a battery.

Find the current flowing across R1. 2

Find the currents flowing across R2 and R3 in the circuit shown in Figure 6.6.

3 Find the voltage drop caused by the total current across R1. 4

1000 m of wire with a cross section equal to 8 mm2 is subjected to 100 V. A current of 4 A is measured across this wire. What is the wire resistivity?

Solutions

Solutions 1

First, we need to calculate the circuit’s total resistance. Resistors R2 and R3 are in parallel. Their equivalent resistance is 1 1 1 = + REQ R2 R3 1 1 1 = + REQ 10000 22000

REQ = 6875 Ω This equivalent resistance is in series with R1. We must sum both to obtain the total resistance: RT = R1 + REQ RT = 1200 + 6875

RT = 8075 Ω Now that we have the total resistance, we can use Ohm’s law and find the current across R1: V = RI 10 = 8075 I I=

10 8075

FINAL RESULT I = 1 2384 mA

2

Current flowing across R1 will split between R2 and R3. Thus, the sum of currents across R2 and R3 will be equal to the total current flowing across R1: I1 = I2 + I3 If R2 is 10 kΩ and R3 is 22 kΩ, we conclude that R3 is 2.2 times bigger than R2, or R3 = 2 2R2 Hence, current in R2 will be 2.2 times bigger than the current flowing across R3:

59

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6 Ohm’s Laws

I2 = 2 2 I3 Upon substituting the values, we get I 1 = 2 2 I3 + I3 I1 = 3 3 I3 Thus, 0 0012384 = 3 3 I3

FINAL RESULT I3 = 375 2697 μA

Now that we know I3, we can find I2: I2 = 2 2I3

FINAL RESULT I2 = 825 5933 μA

3

Current across R1 causes a voltage drop of VR1 = R1 I1 VR1 = 1200 0 0012384

FINAL RESULT VR1 = 1 4860 V

4

We first apply the first Ohm’s law to find the conductor’s resistance: V = RI Thus, R=

V I

Solutions R=

100 4

R = 25 Ω The second Ohm’s law states that the conductor’s resistance is R=

ρL A

ρ=

RA L

or,

Upon substituting the values, we obtain R = 25 Ω A = 8 mm2 = 8 × 10 − 6 m2 L = 1000 m Thus, ρ=

25 × 8 × 10 − 6 = 2 × 10 −7 1000

FINAL RESULT ρ = 200 nΩ m

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63

7 Delta–Wye Conversions Circuit Analysis

7.1

Introduction

In this chapter, we will examine two kinds of circuit configurations, Delta and Wye (Y), and how to convert from one to the other.1 Converting a circuit between Delta and Y configurations and vice versa can make a circuit easier to understand and analyze.

7.2

Delta Circuit

Figures 7.1 and 7.2 show examples of Delta and Y circuit configurations.

7.3

Delta–Wye Conversion

Consider that we want to find the resistance between points A and B in the circuit shown in Figure 7.1 and that point C is not connected to anything. If point C is not connected to anything, we can redraw the circuit into a friendlier version shown in Figure 7.3. Thus, the resistance between A and B will be the resistance of R1 in parallel with the sum of R2 and R3, because these later resistors are in series.

Figure 7.1 Delta or triangle circuit.

Figure 7.2 Y or Wye circuit. 1 Delta configuration is also known as triangular or π and Y configuration is also known as T or Wye. Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

64

7 Delta–Wye Conversions

Mathematically speaking, RAB = R1

R2 + R3

By applying the equations for the equivalent resistance of parallel resistors, Figure 7.3 Delta circuit modified.

RAB =

R1 × R2 + R3 R1 + R2 + R3

If we want to convert the original Delta circuit into a Wye circuit, we must look at the Wye circuit in Figure 7.2 and observe what resistance would be seen between A and B if point C was not connected to anything. The answer is obvious: the sum of RA and RB, or RAB = RA + RB We have now two equations that calculate the resistance between points A and B when C is not connected to anything: one for the Delta circuit and one for the Wye circuit. We can equate both equations for nodes A and B: R1 × R2 + R3 = RA + RB R1 + R2 + R3 If we do the same for the other Delta nodes, we obtain two more equations. For nodes A and C, R2 × R1 + R3 = RA + RC R2 + R1 + R3 and for nodes B and C, R3 × R1 + R2 = RB + RC R3 + R1 + R3 If we solve these three equations, we obtain the equations that can be used to convert any Delta circuit into a Wye circuit. DELTA–WYE CONVERSION EQUATIONS R 1 R2 R 1 + R2 + R3 R1 R3 RB = R1 + R2 + R3 R2 R 3 RC = R1 + R2 + R3

RA =

RA , RB , RC are the resistors in the Wye configuration, in Ohms. R1 , R2 ,R3 are the resistors in the Delta configuration, in Ohms.

7.5 Examples

7.4

Wye–Delta Conversion

If we want to convert Wye circuits to Delta circuits, we must follow the same kind of principles we have used to convert from Delta to Wye. If we do, we find the following equations that can be used to convert any Wye circuit into a Delta one. Y-DELTA CONVERSION EQUATIONS RA RB + RB RC + RA RC RC RA RB + RB RC + RA RC R2 = RB RA RB + RB RC + RA RC R3 = RA

R1 =

RA , RB , RC are the resistors in the Wye configuration, in Ohms. R1 , R2 , R3 are the resistors in the Delta configuration, in Ohms.

If we superimpose the Delta and the Wye circuits, we notice that a new node was created by the conversion from Delta to Wye, in the middle of the diagram, as shown in Figure 7.4. In the next example, it will become clear how useful is the ability to convert a circuit from Delta to Wye or vice versa and how this can help to simplify circuits in real life.

7.5

Examples

7.5.1

Example 1

Figure 7.4 A new node is created after a Delta–Wye conversion.

Find the resistance measured between points A and B in the circuit shown in Figure 7.5. 7.5.1.1

Solution

Finding the resistance between A and B is not easy as it appears. The circuit has sev- Figure 7.5 Delta–Wye conversion. eral resistors that are in series or parallel, and there is nothing obvious that can be replaced by an equivalent resistance. We could apply Ohm’s law, for example, to discover the voltages and currents across the components, but we would end with a lot of equations to solve.

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7 Delta–Wye Conversions

Things will get easy if we use the Delta–Wye conversion equations. To see how this circuit can be converted into a Delta configuration, we stretch the diagram and redraw it like shown in Figure 7.6. It is now clear that the circuit in Figure 7.5 is, in fact, a double Delta configuration, sharing resistor R3. To simplify the circuit, let us replace R1, R3, and R4 by the Y configuration shown superimposed in Figure 7.7. According to the Delta–Wye conversion rules, R1, R3, and R4 must be replaced by Ra, Rb, and Rc, and their values can be calculated by the following equations: Ra = Ra =

R1 R3 R1 + R3 + R4

4 × 10 4 + 10 + 6

Ra = 2 Ω

Rb = Rb =

R3 R4 R1 + R3 + R4

10 × 6 4 + 10 + 6

Rb = 3 Ω

Rc = Rc =

R1 R4 R1 + R3 + R4

4×6 10 + 4 + 6

Figure 7.6 Stretching the circuit to see the Delta.

Figure 7.7 Wye superimposed into de Delta.

7.5 Examples

Rc = 1 2 Ω Now that we have the values, let us rewrite the final circuit as shown in Figure 7.8. Circuit in Figure 7.8 can be redrawn into the one in Figure 7.9. Now we can clearly see that Rc is in series with two branches of resistors in parallel and that each branch has two resistors in series, Ra/R2 and Rb/R5. Mathematically speaking, the resistance between A and B is RAB = Rc + Ra + R2 RAB = Rc +

Figure 7.8 Simplified circuit.

Rb + R5

Ra + R2 × Rb + R5 Ra + R2 + Rb + R5

Figure 7.9 Final simplified circuit.

2+2 × 3+8 2+2 + 3+8

RAB = 1 2 +

RAB = 1 2 +

4 × 11 4 + 11

RAB = 1 2 +

44 15

FINAL RESULT RAB = 4 133 Ω

7.5.2

Example 2

Figure 7.10 shows what appears to be a Delta circuit. We want to find the resistance between points A and B. 7.5.2.1

Solution

The circuit in Figure 7.10 appears to show that R3, R4, and R5 are in a Delta configuration. This is not true. This circuit is what we call a fake Delta circuit, an impostor, because these resistors are not in Delta.

Figure 7.10 Peculiar Delta circuit.

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7 Delta–Wye Conversions

The circuit in Figure 7.10 can be redrawn as the one shown in Figure 7.11. We can improve the circuit more by redrawing it like it is shown in Figure 7.12. Figure 7.12 shows that R1/R3 and R2/R6 are in parallel and both groups are in series. There is no resistor in Delta. The resistance between A and B can be then calculated as RAB = REQ

R5

R6

Figure 7.11 Fake Delta circuit

where REQ is REQ = R1 R3 + R2 R4 R1 × R3 R2 × R4 + REQ = R1 + R3 R2 + R4 REQ =

4 × 10 2 × 6 + 4 + 10 2 + 6

REQ =

40 12 + 14 8

REQ =

61 Ω 14

Hence, the resistance between A and B is 1 1 1 1 = + + RAB REQ R5 R6 1 1 1 1 = + + RAB 61 8 10 14 1 14 1 1 = + + RAB 61 8 10 1 1109 = RAB 2440

FINAL RESULT RAB = 2 2 Ω

Figure 7.12 Peculiar Delta circuit decomposed.

Solutions

Exercises 1

Find the resistance between A and B in the circuit shown in Figure 7.13.

2

Find the resistance between A and B in the circuit shown in Figure 7.14. Figure 7.13 Delta circuit (Exercise 1).

Solutions 1

We can convert from Delta to Wye by replacing R1, R3 and R4 with Ra, Rb and Rc, redrawing the circuit as a Wye configuration, and using the following equations: Ra =

Ra =

R1 R3 R1 + R3 + R4

10 × 4 10 + 4 + 8

Figure 7.14 Delta circuit (Exercise 2).

Ra = 1 8182 Ω

Rb =

Rb =

R3 R4 R1 + R3 + R4

4×8 10 + 4 + 8

Rb = 1 4545 Ω

Rc =

RC =

R1 R4 R1 + R3 + R4

10 × 8 10 + 4 + 8

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7 Delta–Wye Conversions

Rc = 3 6364 Ω This operation will convert the original circuit into the one shown in Figure 7.15. The new circuit is now highly Figure 7.15 Converted circuit (exercise). simplified. Resistors Ra/R2 and Rb/R5 are in series. Both groups are in parallel and the block is in series with Rc. Mathematically speaking, the resistance between A and B can be described by the following equations: RAB = Rc + Ra + R2 RAB = Rc + RAB = 3 6364 +

Rb + R5

Ra + R2 × Rb + R5 Ra + R2 + Rb + R5 1 8182 + 20 × 1 4545 + 12 1 8182 + 20 + 1 4545 + 12

FINAL RESULT RAB = 11 9588 Ω

2

The circuit shown in Figure 7.14 is a Delta impostor and can be redrawn as shown in Figure 7.16. Thus, the resistance between points A and B can be found by the following formula: RAB = REQ

R5

R6

where REQ = R1

R3 + R2

R4

Thus, REQ =

REQ =

R1 × R3 R2 × R4 + R1 + R3 R2 + R4 8×6 4×2 + 8+6 4+2

Figure 7.16 Fake Delta circuit (exercise).

Solutions

REQ = 4 7619 Ω Hence, the resistance between A and B is RAB = REQ

R5

R6

1 1 1 1 = + + RAB REQ R5 R6 1 1 1 1 = + + RAB 4 7619 10 10

FINAL RESULT RAB = 2 4390 Ω

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8 Capacitors And Electric Charges

8.1

Introduction

In this chapter, we will examine capacitors, basic electronic components very popular in all kinds of circuits.

8.2

History

The capacitor or originally known as condenser1 is a two-terminal electric component that can store potential energy in the form of electric field. Technically, a capacitor is composed of two metal plates, separated by a medium. This medium, called dielectric, can be any nonconductive element like air, oil, plastic, etc.

Historically the idea for the capacitor is based on the Leyden Jar.2

8.3

How It Works

Capacitors are basically two metal plates in parallel. The plates are put very close to each other, without touching. These plates are at rest and have roughly the same number of electrons (see Figure 8.1). 1 The term condenser used to refer to capacitors was first used around 1782. It is interesting to know that, in that year, James Watt had just patented a new revolutionary component for his steam engine, called the “separate condenser,” making the term a success among scientists. 2 Leyden Jar (or Leiden Jar) is a kind of primitive capacitor that can store high voltage electric charges between two electric conductors. Invented by accident in 1746 by Pieter van Musschenbroek. Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

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Figure 8.1 Two metallic plates in parallel, at rest.

Something magical happens when a battery is connected to the plates. Suddenly, an electric field (F) is formed between the plates, as shown in Figure 8.2.

Figure 8.2 A battery is connected between the plates.

The electric field makes the plate connected to the battery’s negative pole (A) to accumulate an excess of negative charges (electrons) and the other one to accumulate positive charges, or lack of electrons (B). Because both plates are very close and charges with the same polarity repel each other, this excess of electrons in plate A will create a force that will repel electrons on plate B, expelling them from the plate and forcing them to migrate to the battery’s positive pole.

8.3 How It Works

At this time, one may have the illusion of current flow between the plates. The truth is that the electric field generated by plate A forced electrons from plate B to move off the plate to the battery’s positive pole. No electron crossed through the dielectric from plate A to B.

A real current may flow across the dielectric in some cases, but this is an undesired effect that may destruct the capacitor.

As plate A receives more electrons, it saturates and an increasing repulsion force develops to prevent more electrons from arriving at plate A. The force between plates stops increasing and cannot remove electrons from the other plate anymore. At this point, current stops “flowing.”

Charging a capacitor is like compressing a coil to its physical limit. At some point it is not possible to compress the coil anymore.

Capacitors are normally represented on schematics by the uppercase letter C. Figure 8.3 shows the symbols used to represent a capacitor in the American and in the international standards.

Figure 8.3 Capacitor symbols.

8.3.1

Dielectric

The kind of capacitor we are examining so far is composed of two metallic plates separated by air. Air is the dielectric. We know that air is not a very good conductor. In fact, air is an insulator. What makes a material a good insulator is having very few free electrons. Electrons are charge carriers. A small number of free electrons make it hard to conduct electricity, because all available electrons are held in place by the nuclear strong force, making very hard for them to move. A lot of energy would be required to strip these electrons from their positions and make them flow. Therefore, we conclude that given enough energy an insulator can be transformed into a conductor.

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8 Capacitors

Air, for example, is an insulator and a lightning is an example of how current can flow through an insulator if a colossal energy is provided. Air insulation is about 3 million volts per meter.

A lightning measures, in average, about an inch wide (2.5 cm) and 5 miles (8 km) long and can reach 200 million volts. The most impressive lightning ever recorded occurred in Dallas and crossed an impressive 118 miles (190 km).

A unity called permittivity, measured in Farads per meter (F/m) and referred by the symbol εr or κ, denotes how much the molecules of a material oppose an external electric field. A second unity called relative permittivity, also known as dielectric constant, shows how this permittivity relates to the permittivity of the vacuum. No current will flow across the dielectric until the electric field is strong enough to break it. If so, little “lightnings” or sparks will fly across the plates and current will flow. This is not a desired effect and will cause damage to the capacitor. The voltage at which the dielectric breaks is called the breakdown voltage. In real life, capacitors that can work with very high voltages are needed and air is not enough insulator for these cases. Thus, new dielectrics and construction methods had to be developed. 8.3.2

Construction Methods

Capacitors with simply round or rectangle plates parallel to each other are not practical. They are generally huge in size and have low capacitances. One method for building better capacitors is using long two thin strips of metal foil (A and C) with dielectric material sandwiched between them (B) – everything wounded into a tight roll and sealed in paper or metal tubes – as shown in Figure 8.4. Figure 8.4 Rolled-up capacitors.

8.4 Electric Characteristics

This method, in combination with diverse dielectrics, allows building high capacitance capacitors in a small package. Different dielectrics also make possible to create capacitors with large capacitances, large breakdown voltages, and better performances under certain circumstances, as high frequencies, for example. Table 8.1 shows several materials and their permittivities. Table 8.1 Permittivity. Material

Permittivity (F/m)

Vacuum

1

Glass

3.7–10

Teflon

2.1

Polyurethane

2.25

Polyamide

3.4

Polypropylene

2.2–2.36

Polystyrene

2.4–2.7

Titanium dioxide

86–173

Strontium titanate

310

Strontium barium titanate

500

Barium titanate

1250–10,000

Diverse polymers

1.8–10,000

Calcium copper titanate

> 250,000

8.4

Electric Characteristics

These are the main characteristics of capacitors:

•• ••

They store electric charges in the form of an electric field. An ideal capacitor can store an electric charge indefinitely. In real life, they discharge slowly through the dielectric (leakage), because dielectrics are not perfect insulators and let a little bit of current flow. Capacitors block direct current. Capacitors let alternating current (AC) pass.

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8 Capacitors

Like resistors, capacitors are passive components, because they do not have an active role inside a circuit.

8.5 Electric Field Considering that the plates of a capacitor have an area equal to A and are separated by a distance d, the electric field between the plates can be found by using the following equation. ELECTRIC FIELD E=

qd εA

E is the electric field, in Newtons per Coulomb or volts per meter. q is the amount of stored charge, in Coulombs. ε is the dielectric permittivity, in Farads per meter (F/m). A is the plate area, in squared meters. d is the distance between the plates, in meters.

8.6 Capacitance Capacitance is the ratio of the change in an electric charge in a system to the corresponding change in its electric potential. CAPACITANCE C=

q V

C is the capacitance, in Farads. q is the amount of charge, in Coulombs. V is the electric potential, in Volts.

Capacitance is measured in Farads3 and represented by the uppercase letter F in the SI.

3 In honor of Michael Faraday (1791–1867), an English scientist that pioneered the first studies about electromagnetism and electrochemistry.

8.7 Stored Energy

According to this formula, a capacitor of 1 F can store a charge of 1 C when subjected to 1 V.

There are two kinds of capacitance: one produced by the object itself and one produced by proximity to other objects. Any object that can be electrically charged will have its own value of capacitance. Also, any object in the proximity of others will generate mutual capacitance. Every component in a circuit has capacitance. Most of the time this is a parasitic characteristic.

Capacitance can also be expressed by the following formula. CAPACITANCE C=

εA d

C is the capacitance, in Farads. ε is the dielectric permittivity, in Farads per meter (F/m). A is the plate’s area, in squared meters. d is the distance between the plates, in meters.

8.7

Stored Energy

Capacitors can store a large amount of energy in the form of electric field in a relatively short amount of time, differently from batteries that take a long time to charge. However, capacitors cannot supply energy or hold their electric charge for a long time because they slowly discharge through their own dielectrics or through the circuit they are connected. However, their charge and discharge characteristics can be utilized to create very interesting circuits, like audio filters, blinkers, and timers. Capacitors are frequently used to stabilize voltage levels on power supplies.

The energy stored inside a capacitor can be expressed in terms of work to move charges from a plate to another. In other words, the energy stored inside a capacitor is equivalent to the work necessary to keep positive and negative charges, +q and −q, in their respective plates.

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8 Capacitors

Mathematically, this work and its equivalent charge can be related by the following equation. STORED ENERGY OR WORK dW =

q dq C

W is the work, in Joules. q is the stored charge, in Coulombs. C is the capacitance, in Farads.

To find the total energy, we can integrate the previous equation: W=

1 C

Q

q dq 0

MATH CONCEPT

f x dx

The integral of

x +C 2

The constant of integration (C) can be ignored when dealing with defined integrals. See Appendix F.

Hence, 1 q2 W= × C 2

W=

Q dq 0

1 Q2 02 − × 2 2 C

TOTAL ENERGY W=

Q2 C

W is the total energy, in Joules. Q is the total charge, in Coulombs. C is the capacitance, in Farads.

8.8 Voltage and Current

We know that Q = CV, and thus we can convert the previous formula into the following equation. TOTAL ENERGY W=

CV2 2

W is the total energy, in Joules. V is the voltage, in Volts. C is the capacitance, in Farads.

8.8

Voltage and Current

Like mentioned before, when a discharged capacitor is connected to a continuous voltage source, such as a battery, one of its plates will be filled with an excess of charges and the other will have a lack of charges. This excess of charges in one of the plates forms a strong electric field that expels electrons from the other plate and makes them move to the battery. Over time, the first negative plate will become saturated with charges and the process will stop. For all intents and purposes, we can say that a current flowed through the capacitor. 8.8.1

Current on a Charging Capacitor

During the charge process of an initially discharged capacitor, the charging current will start at the maximum value possible and will exponentially decay to 0 as the capacitor charges. Figure 8.5 shows how the current behaves during charge. Figure 8.5 Current on a charging capacitor.

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8 Capacitors

Current across a capacitor follows the following equation. CURRENT ON A CHARGING CAPACITOR

i=C

dv dt

i is the current, in Amperes. C is the capacitance, in Farads. dv/dt is the ratio of voltage change over time.

The previous equation tells us that current in a capacitor is proportional to the ratio between voltage variations over time. We know that charge is also equal to Q = CV. Therefore, if we derive both sides, dQ dV =C dt dt Upon substituting this relation in the previous formula, we obtain the following equation. CURRENT ACROSS A CAPACITOR

i=

dQ dt

i is the current, in Amperes. dQ/dt is the ratio of charge change over time.

8.8.2

Voltage on a Charging Capacitor

During the charge process of an initially discharged capacitor, voltage will start at 0 and increase exponentially until it reaches the charging level. As the voltage across the capacitor approaches the charging one, the charging speed will decrease until it reaches 0 and the capacitor is fully charged. Figure 8.6 shows how the voltage across a capacitor behaves over time in a charging capacitor.

8.8 Voltage and Current

Figure 8.6 Voltage on a charging capacitor.

Rearranging the previous equation, we see that the voltage across a capacitor can be written as i=C

dv dt

dv i = dt C dv =

i dt C

We can integrate both sides and obtain the equation of voltage across a capacitor during charge. VOLTAGE ACROSS A CAPACITOR

v t =

1 i t dt C

v(t) is the equation of voltage across a capacitor as a function of time, in Volts. C is the capacitance, in Farads. i(t) is the equation of current as a function of time, in Amperes.

If we integrate for a specific interval, we get the following equation.

83

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8 Capacitors

VOLTAGE ACROSS A CAPACITOR v t =

1 C

T

i t dt + V0

81

0

v(t) is the equation of voltage across a capacitor as a function of time, in Volts. C is the capacitance, in Farads. i(t) is the equation of current as a function of time, in Amperes. T is the considered time, in seconds. V0 is the capacitor’s initial voltage, in Volts.

8.9 Examples 8.9.1

Example 1

Suppose we apply a current of 5 mA for 4 ms across an ideal 10 μF capacitor. The capacitor is initially discharged. We will call this sudden current a “current pulse.” What is the voltage across the capacitor before and after the current pulse? 8.9.1.1 Solution

We know that integrals always calculate the area below the functions they represent. If we want to find the voltage across a capacitor at a specific instant in time, we must consider the equation for the voltage across the capacitor during charge: 1 T i t dt + V0 C 0 This equation describes voltage in terms of current as a function of time. In other words, this equation describes the current’s area for a specific interval. For this example, the area is the one shown on Figure 8.7. v t =

Figure 8.7 Current area.

8.9 Examples

8.9.1.2

Before the Pulse

Before the pulse, the capacitor is completely discharged. Hence, its voltage is 0. 8.9.1.3

During the Pulse

During the pulse, the capacitor charges exponentially and therefore its voltage increases. Voltage across a capacitor is ruled by the following equation: v t =

T

1 C

i t dt + V0 0

The current pulse we have applied is constant. Hence, we can substitute the function i(t) by a fixed value i and remove it from the integral. Also, V0 is equal to 0, because the capacitor is initially discharged. Upon substituting these values, we get v t =

T

1 i C

dt 0

MATH CONCEPT

dx

The integral of

x+C

The constant of integration (C) can be ignored for defined integrals. See Appendix F.

Solving the integral, i v t = t C

t 0

i T −0 C i v= T C Upon substituting the values in the formula, we obtain the voltage across the capacitor after 4 ms: v=

i = 5 mA C = 10 μF t = 4 ms

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8 Capacitors

v=

i T C

v 4 ms =

5 × 10 − 3 4 × 10 −3 10 × 10 −6

FINAL RESULT v 4 ms = 2 V

8.9.1.4 After the Pulse

After the pulse, current drops to 0 and the capacitor stops charging. An ideal capacitor will never discharge and hold its charge indefinitely. Consequently, the capacitor will keep its voltage equal to 2 V forever. 8.9.2

Example 2

We repeat the same example as before but using a 1000 μF capacitor. What is the voltage across the capacitor after the same current pulse is applied? 8.9.2.1 Solution

Upon substituting the values in the formula, we obtain the voltage across the capacitor after 4 ms: i = 5 mA C = 1000 μF t = 4 ms v=

i T C

v 4 ms =

5 × 10 −3 4 × 10 −3 1000 × 10 − 6

FINAL RESULT v 4 ms = 20 mV

8.10 AC Analysis

For the same pulse, the voltage across the capacitor is very small now. This is expected because the capacitance is bigger and, for that reason, the capacitor takes more time to charge.

8.10

AC Analysis

We will examine, now, what happens when a capacitor is connected to an AC source. Capacitors have the property to block direct current and let pass AC under certain conditions. Capacitors react when subjected to AC by offering a resistance to the current flow. This resistance, known as reactance, varies with the wave’s frequency. This property can be used to create filters and manipulate, for example, audio4 signals. 8.10.1

Pool Effect

Some of the properties of capacitors can be explained by an analogy with pools. Imagine an empty pool with two pipes. One of the pipes feeds the pool with water and the other drains the water out. The input pipe is larger and is always closed. It opens, when necessary, to let water in. The output pipe is thin and is always open, letting water out, continuously. If the pool is empty and we let water in, it will take a while to fill. The time it takes to fill depends on how much water enters the pool per second and the size of the pool. If the pool is full and we close the input pipe, the pool will continue to hold its level of water for a while but the water level will decrease slowly, because the output pipe is draining the pool. If the pool is half full and the water flow at the input pipe starts to fail or variate dramatically, the output flux will not be affected and continue to flow as before. We conclude that the pool creates a kind of buffer or cushion effect, keeping the output flow stabilized. Looking through another point of view, we can say that the pool resists drastic changes in the input flow of water, a kind of inertia that will “smooth” drastic changes. Capacitors have the same behavior. As soon as a voltage is applied, they will start to charge. If the voltage increases beyond the capacitor current level, it will charge to match the input level. If the input voltage decreases, the capacitor will discharge to match the input level. If the input voltage is removed, the capacitor will keep its charge for a while but will slowly discharge.

4 Yes, audio is an alternating voltage/current.

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Sudden and drastic variations of the input voltage will be softened and damped, and the voltage across the capacitor will remain relatively constant.

8.11

Capacitive Reactance

Capacitors resist to any sudden changes in voltage. This resistance is called capacitive reactance, represented, normally, by the symbol Xc and measured in Ohms (Ω). Capacitive reactance is represented by the following formula. CAPACITIVE REACTANCE XC =

1 2πfC

XC is the capacitive reactance, in Ohms. f is input’s frequency, in Hertz. C is the capacitance, in Farads. or by

XC =

1 ωC

considering that ω = 2πf. Capacitive reactance and frequency are inversely proportional; if frequency increases, the reactance decreases. If frequency reduces gradually toward 0,5 capacitive reactance will skyrocket to infinity. This explains why capacitors block direct current.6

8.12

Phase

Like explained before, capacitors resist sudden changes in voltage. This resistance represents a kind of inertia that prevents capacitors from reacting instantly to voltage changes. When a capacitor starts to charge, current will be maximum and voltage minimum. When the capacitor is charged, voltage will be maximum and current will be 0.

5 Remember limits. 6 Direct current is a kind of alternating current with a frequency equal to 0.

8.12 Phase

It is clear the existence of a delay between voltage and current and that voltage is 90 behind current. Technically speaking, voltage and current are 90 out of phase. 8.12.1

Mathematical Proof

Like examined before, the voltage across a capacitor can be described in terms of current by this formula: v t =

1 C

T

i t dt

82

0

If we apply an alternated current across a capacitor, this current follows a sinusoidal form: I = IP sin ωt

83

where IP is the peak current and I is the instantaneous current. Substituting (8.3) in (8.2), we get v t =

T

1 Ip C

sin ωt dt

0

To solve this integral we must use the substitution rule. Let u = ωt Thus, du = ωdt Hence, du = dt ω Substituting this on the integral, v t =

Ip C

Ip v t = ωC

T

sin u 0

du ω

T

sin u du 0

MATH CONCEPT The integral of sin(x)dx − cos (x)

84

89

90

8 Capacitors

By applying the previous concept, we get v t =

IP ωC

−cos u

MATH CONCEPT The integral of cos x dx sin θ − π2

By applying this other concept, we obtain the final equation for the voltage across a capacitor. VOLTAGE ACROSS A CAPACITOR v t =

IP π sin ωt − ωC 2

85

v(t) is the equation for the voltage across the capacitor, in Volts. IP is the current’s peak current, in Amperes. ω is the angular frequency, in radians per second, equal to 2πf. C is the capacitance, in Farads. t is the instant in time, in seconds.

What do we see when we compare equations (8.3) and (8.5)? We see that the first equation describes current in terms of sin(ωt) and the second describes the voltage in terms of sin(ωt − π/2). The voltage equation is telling us to subtract π/2, equivalent to 90 , to the angle, meaning that the voltage is delayed 90 in relation to current. Figure 8.8 shows a diagram of voltage curve (second curve) that is 90 behind the current curve (first curve). Figure 8.8 Voltage (in black) is behind current (in red).

8.13 Electrolytic Capacitor

In real life, the angle between current and voltage angles may not be exactly equal to 90 and may vary, due to imperfections and different capacitance values, but it will always be less than 90 .

8.13

Electrolytic Capacitor

To increase the capacitance of a capacitor, we can do a few things:

• • •

Increase the plate areas, but this is not practical and will make capacitors huge. Create new dielectrics that can increase the capacitance by increasing the insulation between the plates. Reduce the distance between the plates, but this is a problem, because the distance between the plates depends on how thin we can make the dielectric.

To overcome these problems, a new kind of capacitor was invented, called electrolyte capacitor, where the plates are formed by an aluminum foil with an oxide layer, a tissue soaked in electrolyte liquid, and a regular aluminum foil. The oxide layer on the first aluminum foil insulates it from the electrolyte liquid, becoming the dielectric and making the electrolyte liquid itself the other “plate,” a kind of virtual plate. For that reason, the electrolyte capacitor is not formed by two metal plates, but rather by an oxidized metal plate and an electrolyte element. The oxidized layer serves as a dielectric and the second regular aluminum foil serves just to supply voltage to the electrolyte. Therefore, an electrolyte capacitor is composed of a metal foil (A) with an aluminum oxide layer (B), a tissue soaked in electrolyte liquid (C), and a regular metal foil (D), as shown in Figure 8.9. Figure 8.9 Several layers of an electrolyte capacitor (cross section).

91

92

8 Capacitors

The electrolyte capacitor is far from being a traditional capacitor. In fact, the process that creates the oxide layer on one of the plates makes the electrolyte capacitor only able to conduct current in one direction. If an electrolyte capacitor is subjected to a reverse current flow, it will be destroyed and probably explode. Therefore, these electrolyte capacitors, also known as “polarized capacitors,” have a minus sign on their bodies to identify their polarity. The advantages of electrolytic capacitors are as follows:

• • ••

Their dielectric layer is very thin, allowing the plates to be very close, thus increasing the capacitance. They are extremely tough, have a high degree of insulation, and can heal themselves up to a certain degree during their lifespan. High capacitance electrolyte capacitors have very small sizes. Very cheap to produce.

Figure 8.10 shows a representation of an electrolytic capacitor where a huge minus sign can be seen on the stripe and an X is seen on the metallic end.

Figure 8.10 A radial electrolyte capacitor.

This X at ending is, in fact, a groove to create a weakened part of the casing, which will bulge upward and help the capacitor release its contents in case of failure. If so, they will not explode violently hurting people or causing damage to the whole equipment. These grooves are designed to prevent catastrophic failures from happening. Figure 8.11 shows the symbols for the electrolytic capacitor in the United States (first symbol) and in the world (last two symbols).

Figure 8.11 Electrolyte capacitor symbols in the United States and internationally.

8.15 Capacitors in Series

8.14

Variable Capacitors

Some applications in the real world require capacitors that can have a variable capacitance. For that reason, a new capacitor was born, the variable capacitors or trimmers. The first line in Figure 8.12 shows the symbols for the variable capacitors used in the United States, and the second line shows the symbols used internationally. Figure 8.12 Variable capacitors.

8.15

Capacitors in Series

If we connect two capacitors in series, like shown in Figure 8.13, the equivalent capacitance between points A and B can be found by using the following formula: 1 1 1 = + C C1 C2

Figure 8.13 Capacitors in series.

93

94

8 Capacitors

For an infinite number of capacitors, a more generic formula must be used. SERIES CAPACITANCE 1 1 1 + + = C C1 C2

+

1 C∞

C is the total equivalent capacitance, in Farads. C1 , C2 , … is the capacitance of each individual capacitor, in Farads. If there are just two capacitors in series, a friendlier formula can be used:

C=

8.16

C1 × C2 C1 + C2

Capacitors in Parallel

If we connect two capacitors in parallel, like shown in Figure 8.14, the equivalent capacitance between points A and B can be found by using the following formula: C = C1 + C2 For an infinite number of capacitors, a more generic formula must be used. PARALLEL CAPACITANCE C = C1 + C2 +

+ C∞

C is the total equivalent capacitance, in Farads. C1 , C2 , … is the capacitance of each individual capacitor, in Farads.

Figure 8.14 Capacitors in parallel.

8.17 Capacitor Color Code

8.17

Capacitor Color Code

In the past, capacitors, as well as resistors, were identified by a color code painted on their bodies. Today this is not the case anymore. However, it is worth to know how this code works because it is not rare to find capacitors like that on old equipment. 8.17.1

The Code

The code was composed of four or five stripes painted on their bodies. The capacitance is represented by the four-color code: the first two stripes represent the first digits, the third stripe is the multiplier, and the last stripe is the maximum operational voltage. In the five-color code, the first two stripes represent the first two digits, the third stripe is the multiplier, the fourth stripe represents the tolerance, and the last one represents maximum operational voltage (Figure 8.15).

Figure 8.15 Four- and five-stripe capacitors.

Table 8.2 shows the values for each stripe in both codes.

The color code result is always expressed in picofarads.

A capacitor with the stripes shown next will have the following value: Brown

1

Black

0

Orange

×1000

Brown

±1%

The first three colors mean 10 × 1000 pF or 10000 pF, which is equivalent to 10 nF.

95

96

8 Capacitors

Table 8.2 Capacitor color code.

Color

First Stripe

Second Stripe

Multiplier

Tolerance >10 pF (%)

Tolerance ω0, equivalent to an overdamped circuit. The equation for the voltage across the capacitor, for this case, is VC t = Ae m1 t + Be m2 t + V1

27 39

Solutions

The roots of an overdamped solution are m1 = − α +

α2 − ω20

m1 = −25 +

252 −1 22472

m1 = − 0 03 and m2 = −25−

252 − 1 22472

m2 = − 49 97 The coefficients in Eq. (27.37) are A=

V0 −V1 m1 1− m2

0− 12 − 0 03 1− −49 97

A=

A = − 12 0072 and B=

B=

V0 −V1 m2 1− m1

0− 12 −49 997 1− − 0 03

B = 0 0072 Having found all unknowns, we can write the final equation for the voltage across the capacitor (27.39) as VC t = Ae m1 t + Be m2 t + V1

421

422

27 RLC Circuits: Part 1

FINAL RESULT VC t = −12 0072e − 0 03t + 0 0072e −49 97t + 12

What is the voltage across the capacitor 5s after the switch is closed? We use the last equation and solve for t = 5 s: VC 5 s = −12 0072e −0 03 5 + 0 0072e − 49 97 5 + 12

FINAL RESULT VC 5 s = 1 6653 V

2

The complete equation for the voltage across the capacitor. The given circuit is ruled by the following equation: 2

d2 VC dVC + 40 VC = 0 +8 dt2 dt

We divide the equation by 2 to write it in the standard form as d2 V C dVC + 20 VC = 0 +4 dt2 dt

27 40

Series RLC circuits are ruled by the following second-order differential equation: d2 VC R dVC 1 VC = 0 + + L dt LC dt2 which we rewrite as follows: A 2

d Vtr t + dt2

B

R dVtr t 1 + Vtr t = 0 L LC dt

Solutions

If we compare A and B of this equation with the same values of Eq. (27.40), which we rewrite in the following, A

B

4

dVC t + 20 VC t = 0 dt

2

d VC t + dt2

27 41

we can extract several values: R =4 L

27 42

In this given example, R=2 Ω By substituting R in (27.42), we get L=0 5 H Another part we can extract by comparing the equations is 1 = 20 LC

27 43

Therefore, by substituting L in (27.43), we get

C=

1 1 = = 0 1F 20L 20 0 5

The next thing we must do is to find if this is a critically damped, overdamped, or underdamped circuit. We proceed to calculate α α= α=

R 2L

2 2 05

423

424

27 RLC Circuits: Part 1

α=2 and ω0 1 LC

ωo = ωo =

1 05 01

ωo = 4 4721 We see that α < ω0. Therefore, this is an underdamped series RLC circuit, ruled by the following equation for the voltage across the capacitor: VC t = e −αt K1 cos ωd t + K2 sin ωd t + V1 We proceed finding the unknowns: first K1 K1 = V0 −V1 K1 = 0−12

K1 = − 12 and then K2 K2 =

α V 0 − V1 ωd

To find K2 we need to find ωd: ωd =

α2 −ω20

ωd =

22 −4 47212

ωd ≈ j4

Solutions

Therefore, K2 =

2 0− 12 4

K2 = −6 The complete equation for the voltage across the capacitor can be assembled as follows. FINAL RESULT VC t = e −2t − 12 cos 4t − 6 sin 4t + 10

The voltage across the capacitor for t = 3 s. Using the last equation, we can solve for t = 3 s: VC 3 s = e − 2 3 −12 cos 4 3 −6 sin 4 3 VC 3 s = e − 6 −12 cos 12 − 6 sin 12 + 10

FINAL RESULT VC 3 s = 9 9828 V

+ 10

425

427

28 RLC Circuits: Part 2 Current Analysis in Circuits Containing Resistors, Capacitors, and Inductors in Series

28.1

Introduction

In the last chapter, we have performed a voltage analysis in RLC series circuits, finding their voltage equations for various specific cases. In this chapter we will continue to examine series RLC circuits in terms of current analysis.

28.2

The Circuit

Our analysis will focus on the same circuit we have used on the last chapter: a direct voltage source, a resistor, a capacitor, an inductor, and a switch in series, like shown in Figure 28.1. Figure 28.1 A basic RLC circuit.

Like before, the switch is initially open, there is no current flowing in the circuit, the inductor is completely de-energized, and the capacitor is completely discharged. At time t = 0 s, the switch is closed, and current flows and reaches the inductor. Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

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28 RLC Circuits: Part 2

28.2.1

Current

As soon as the switch is closed, we can apply to the circuit. Kirchhoff’s voltage law (KVL) states that the sum of all voltages in a closed loop is 0. Therefore, the power supply voltage (V1) is equal to the voltage drop on the resistor added to the voltage across the inductor and capacitor, or V 1 = VR t + V L t + V C t

28 1

Voltages across a capacitor, an inductor, and a resistor are defined as 1 idt C VR t = Ri t

VC t =

VL t = L

di t dt

By substituting this in (28.1), we get

V1 = Ri t + L

di t 1 + idt C dt

If we take the derivative of both sides of this equation with respect to time, we obtain d d d di t V1 t = R i t + L dt dt dt dt

+

1d C dt

idt

V1(t) is the power supply voltage, a constant. We know that the derivative of a constant is 0; therefore, d V1 t dt 0=R

0

=R

d d di t it + dt dt dt

d d di t it + dt dt dt

+

1d C dt

+

1d C dt

idt

idt

The last term is the derivative of an integral, that is, one thing cancels the other, and we have just the function.

28.2 The Circuit

Therefore, 0=R

d d di t it + dt dt dt

+

1 i C

or after rearranging the equation, we have L

d2 i t di t 1 + i=0 +R C dt2 dt

or written in standard form: d2 i t R di t 1 i=0 + + 2 L dt LC dt

28 2

This second-order differential equation for the current (28.2) is, obviously, very similar to the one described on the previous chapter for voltage across a capacitor. For the same reasons as before, we know that the transient equation for the current will have the form y = Ae mt Consequently, we can convert Eq. (28.2) into d2 Rd 1 Ae mt + Ae mt = 0 Ae mt + 2 L dt LC dt

28 3

The first two terms are derivative of exponentials that we already know how to solve, which will produce the following result for the first term and for the second term: m2 Ae mt and

R m Ae mt L

429

430

28 RLC Circuits: Part 2

By putting these terms back into Eq. (28.3), we get R 1 m2 Ae mt + m Ae mt + Ae mt = 0 L LC R 1 m2 + m + L LC

Ae mt = 0

28 4

Equation (28.4) is the same we have found for the voltage across the capacitor in the previous chapter. Consequently, we know that this equation will have three possible solutions for the current: critically damped, overdamped, and underdamped.

28.3

Current Equations

There is no need to repeat the same calculation we have done in the last chapter. The final equations for the three cases will be the following. 28.3.1

Critically Damped Solution

CURRENT EQUATION – CRITICALLY DAMPED SOLUTION i t = At + B e − αt i(t) is the equation of current as a function of time, in Amperes. A and B are coefficients, to be determined by the circuit’s initial conditions, i(0) and i 0 . α is the damping coefficient. t is the time, in seconds.

28.3.1.1 Current Curve

Figure 28.2 shows the curve for the current flowing in the critically damped series RLC circuit, as soon as the switch is closed. Notice how the current grows from 0 and hits a peak on the positive side and then decays slowly to 0.

Figure 28.2 Current flowing in the circuit (critically damped solution).

This curve was plotted for a critically damped series RLC circuit like the one in Figure 28.1, with a 1H inductor, a 1F capacitor, a 2 Ω resistor, and a power supply V1 of 10 V.

28.3 Current Equations

28.3.2

Overdamped Solution

CURRENT EQUATION – OVERDAMPED SOLUTION i t = Ae m1 t + Be m2 t i(t)is the equation of current as a function of time, in Amperes. A and B are coefficients, to be determined by the circuit’s initial conditions, i(0) and i 0 . t is the time, in seconds. α is the damping coefficient. m1 and m2 are the roots equal to m1 = − α + α2 − ω20 m2 = − α− R α = 2L ωo = 1LC

α2 −ω20

28.3.2.1 Current Curve

Figure 28.3 shows the curve for the current flowing in the overdamped series RLC circuit, as soon as the switch is closed. Notice how the current grows fast from 0 and hits a peak on the positive side and then decays exponentially slowly to 0. This curve was plotted for an overdamped series RLC circuit like the one in Figure 28.1, with a 1H inductor, a 1F capacitor, a 10 Ω resistor, and a power supply V1 of 10 V.

28.3.3

Figure 28.3 Current flowing in the circuit (overdamped solution).

Underdamped Solution

CURRENT EQUATION – UNDERDAMPED SOLUTION i t = e − αt K1 cos ωd t + K2 sin ωd t i(t) is the equation of current as a function of time, in Amperes. K1 and K2 are coefficients, to be determined by the circuit’s initial conditions, i(0) and i 0 . t is the time, in seconds. α is the damping coefficient. R α = 2L ωd =

α2 −ω20

431

432

28 RLC Circuits: Part 2

28.3.3.1 Current Curve

Figure 28.4 shows the curve for the current flowing in the underdamped series RLC circuit, as soon as the switch is closed. Notice how the current oscillates from positive to negative values until it dies. The amplitude decays according to an exponential pattern. Figure 28.4 Current flowing in the circuit (underdamped solution).

28.4

Examples

28.4.1

Example 1

This curve was plotted for a critically damped series RLC circuit like the one in Figure 28.1, with a 1H inductor, a 1F capacitor, a 0.3 Ω resistor, and a power supply V1 of 10 V.

Figure 28.5 shows a circuit with a direct voltage source, a switch, a resistor, an inductor, and a capacitor in series.

Figure 28.5 A basic RLC circuit.

The switch is open, there is no current flowing in the circuit, the inductor is completely de-energized, and the capacitor is completely discharged. At time t = 0 s, the switch is closed, and current flows and reaches the inductor. The following is what we want to find about the circuit:

••

What is the current across the circuit after the switch is closed? What is the current across the circuit two seconds after the switch is closed?

28.4 Examples

28.4.1.1 What Is the Current Across the Circuit After the Switch Is Closed?

When the switch is closed, current flows and reaches the inductor. The inductor hates sudden variations of current and will prevent current from flowing as soon as the switch is closed. FINAL RESULT i 0+ = 0

28.4.1.2 What Is the Current Across the Circuit Two Seconds After the Switch Is Closed?

To find the current equation for this circuit, we must find several parameters and coefficients like α, ω0, and ωd. This circuit is the same used in last chapter, and, therefore, the values are already known as α = 0 005 ω0 = 0 3162 ωd = j0 3161 This is an underdamped circuit because α < ω0; therefore, the current equation has the form i t = e − αt K1 cos ωd t + K2 sin ωd t

28 5

Upon substituting the values, we get i t = e − 0 005t K1 cos 0 31618t + K2 sin 0 31618t

28 6

To find K1 and K2, we must apply the initial conditions for the circuit. 28.4.1.2.1

First Condition: Current at Time Zero

To find the first constant, we must take equation (28.6) and solve for t = 0 s: i t = e −0 005 0 K1 cos 0 31618 0 + K2 sin 0 31618 0 i t = 0 = e0 K1 cos 0 + K2 sin 0 0 = 1 K1 + 0 K2

433

434

28 RLC Circuits: Part 2

K1 = 0

28 7

28.4.1.2.2 Second Condition: The Derivative of the Current Equation for Time Zero

To find the other constant, we take the derivative of equation (28.5) with respect to time and solve for t = 0 s. Therefore, we take the current equation i t = e −αt K1 cos ωd t + K2 sin ωd t and take the derivative d d i t = e −αt K1 cos ωd t + K2 sin ωd t dt dt The first term is the derivative of current with respect to time and we know that the voltage across an inductor is equal, exactly, to the derivative of current with respect to time multiplied by the inductance, or VL = L

di dt

Therefore, di VL = dt L We can substitute this relation in the equation VL d −αt = e K1 cos ωd t + K2 sin ωd t L dt We are before a derivative of a product of functions (u and v) and we can use the product rule: u

v

VL d = e −αt K1 cos ωd t + K2 sin ωd t L dt

28.4 Examples

MATH CONCEPT The derivative of a product of two functions is equal to the derivative of the first function multiplied by the second function added to the first function multiplied by the derivative of the second, or u v = u v + uv

Therefore, the derivatives will expand to VL d − αt e = × K1 cos ωd t + K2 sin ωd t dt L d + e −αt × K1 cos ωd t + K2 sin ωd t dt that can be expanded to

VL d − αt e = × K1 cos ωd t + K2 sin ωd t dt L d d + e − αt × K2 sin ωd t + e −αt × K1 cos ωd t dt dt

28 8

Equation (28.8) has three derivative terms. The first term is the derivative of an exponential and will produce the following result: −αe − αt K1 cos ωd t + K2 sin ωd t

28 9

The second term is the derivative of a cosine. MATH CONCEPT The derivative of d − k sin θ cos kθ dt

Therefore, this produces the following result: −e −αt K1 ωd sin ωd t

28 10

435

436

28 RLC Circuits: Part 2

The third term is the derivative of a sine. MATH CONCEPT The derivative of d k cos θ sin kθ dt

Therefore, this produces the following result: e −αt K2 ωd cos ωd t

28 11

By putting terms (28.9), (28.10), and (28.11) together in Eq. (28.8), we obtain VL = −αe − αt K1 cos ωd t + K2 sin ωd t −e −αt K1 ωd sin ωd t L + e − αt K2 ωd cos ωd t Rewriting VL = −αe − αt K1 cos ωd t + K2 sin ωd t −e −αt K1 ωd sin ωd t L + e − αt K2 ωd cos ωd t Substituting the K1 that we have found in (28.7), we get VL = −αe − αt 0 cos ωd t + K2 sin ωd t −e −αt 0 ωd sin ωd t L + e − αt K2 ωd cos ωd t Because we are taking the derivative at t = 0 s, we can discard also the terms containing sin(ωdt), because sin(0) is zero. Therefore, we obtain

VL = e − αt K2 ωd cos ωd t L As soon as the switch is closed, no current crosses the inductor because it generates an electromotive force to block the flow of current. This electromotive force is equal to the power supply voltage. Therefore, the voltage across the inductor VL, the moment the switch is closed, is equal to V1. Therefore, upon substituting the known values on the equation V1 = e −αt K2 ωd cos ωd t L

28.4 Examples

we get 12 = K2 e −0 005t 0 3161 cos 0 3161t 10 Solving for t = 0 s 12 = K2 e0 0 3161 cos 0 10 12 1 = K2 e0 0 3161 cos 0 10 12 = K2 0 3161 10 1 2 = 0 3161 K2

K2 = 3 7963 We can now assemble the final equation as i t = e −αt K1 cos ωd t + K2 sin ωd t i t = e −0 005t 0 + 3 7963sin 0 3161 t

FINAL RESULT i t = 3 7963 e − 0 005t sin 0 3161 t

The current for t = 2 s is i 2 s = 3 7963 e −0 005 2 sin 0 3161 2

FINAL RESULT i 2 s = 2 2218 A

28.4.2

Example 2

Consider the same circuit shown in Figure 28.5 in the previous example with two changes: the resistor is an 8 Ω, instead of 0.1 Ω, and the power supply is a 10 V direct voltage source, instead of 12 V.

437

438

28 RLC Circuits: Part 2

This new circuit is ruled by the following second-order differential equation:

2

di2 t R di t 1 + i=0 +2 2 L dt 8 dt

Find the equation for the current that rules this circuit. 28.4.2.1 Solution

We divide the circuit’s equation by two to rewrite it using the standard form: di2 t di t 1 + i=0 + 16 dt2 dt Series RLC circuits follow second-order differential equations in the form di2 t R di t 1 i=0 + + L dt LC dt2 By comparing these two equations, we can extract the following information: R =1 L We get L by substituting R = 8 Ω:

L=

8 = 8H 1

We can also extract 1 1 = LC 16 Therefore, we obtain C:

C=

16 16 = = 2F L 8

Using the values of R, L, and C, we can obtain α= α=

R 2L

8 28

28.4 Examples

α=0 5

ωo = ωo =

1 LC 1 82

ωo = 0 25 Because α > ω0, this is an overdamped series RLC circuit, and the current equation follows the form i t = Ae m1 t + Be m2 t

28 12

To find A and B, we apply the initial conditions. 28.4.2.1.1

First Condition: Solve the Current Equation for t = 0 s

We take equation (28.12) and solve for t = 0 s: 0

i 0 = 0 = Ae m1 t + Be m2 t

MATH CONCEPT e0 = 1

0=A+B 0=A+B A = −B

0

439

440

28 RLC Circuits: Part 2

28.4.2.1.2 Second Condition: Take the Derivative of the Current Equation and Solve for t = 0 s

If we take the derivative of the current equation (28.12), we get di t d d = Ae m1 t + Be m2 t dt dt dt

28 13

The first term is the derivative of current equation with respect to time. But we know that the voltage across the inductor is equal to

VL = L

di t dt

28 14

If we rearrange (28.14), we can write it as the derivative of current with respect to time: di VL = dt L which we can substitute in (28.13), making it like VL d d = Ae m1 t + Be m2 t dt L dt The voltage across the inductor, or VL, is the voltage across the inductor immediately after the switch is closed, which is known to be the same voltage as the power supply, as explained before. Therefore, the equation can be converted into 10 d d = Ae m1 t + Be m2 t 8 dt dt d d 1 25 = Ae m1 t + Be m2 t dt dt 1 25 = m1 Ae m1 t + m2 Be m2 t Solving for t = 0 s, 1 25 = m1 Ae0 + m2 Be0 1

1 25 = m1 A e0 + m2 B e0

1 25 = m1 A + m2 B

1

28.4 Examples

We know that A = − B; therefore, 1 25 = m1 − B + m2 B 1 25 = B m2 − m1

B=

1 25 m2 −m1

To find m1 and m2, we must find the equation roots: d2 VC dVC 1 + VC = 0 + 16 dt2 dt which is equal to m2 + m +

1 =0 16

and we can find its roots by using the quadratic equation:

m=

m=

m=

− b ± b2 − 4ac 2a

−1 ±

12 − 4 1 21

−1 ± 0 75 2

Therefore, m1 is m1 =

−1 + 0 75 2

m1 = − 0 0670 and m2 is m2 =

−1− 0 75 2

1 16

441

442

28 RLC Circuits: Part 2

m2 = −0 9330 Therefore, B= B=

1 25 m 2 − m1

1 25 − 0 9330− −0 0670

B = −1 4434 and A will be

A = −B

A = 1 4434 Finally, the equation can be written as follows. FINAL RESULT i t = 1 4434e −0 0670t − 1 4434e −0 9330t

Exercises 1

Figure 28.6 shows a circuit like the one we have been analyzing in this chapter, a resistor, an inductor, a capacitor, and a switch in series with a direct voltage power supply. Figure 28.6 A basic RLC circuit.

Solutions

The switch is initially open, the inductor is completely de-energized, and the capacitor is completely discharged. The circuit’s components have the following values: R = 100 Ω L=2 H C = 800 μF V1 = 30 V At time t = 0 s, the switch is closed, and current flows and reaches the inductor. The following is what we want to find about the circuit:

•• 2

What is the current flowing in the circuit as soon as the switch is closed? What is the current flowing in the circuit 10 ms after the switch is closed?

Consider a series RLC circuit like the previous one. The values of L and C are unknown but the resistor R = 1 Ω and the power supply V1 = 30 V. This circuit is ruled by the following equation:

400

d2 i di + 40 + i = 0 2 dt dt

We know that the current is 0 for t = 0 s. In t = 0 s, the switch is closed. Find the current equation and its value for t = 1 ms.

Solutions 1

What is the current flowing in the circuit as soon as the switch is closed? After the switch is closed, current flows and reaches the inductor. The inductor hates sudden current variations and will not let this new current pass.

FINAL RESULT i 0 + = 0 mA

What is the current flowing in the circuit 5 ms after the switch is closed? We can have three kinds of series RLC circuits, namely, critically damped, overdamped, and underdamped, each one with its own equations for the current.

443

444

28 RLC Circuits: Part 2

The first thing we need to do is to know which circuit is this one. The given values are R = 100 Ω L=2 H C = 800 μF V1 = 30 V We first calculate α and ωo: α= α=

R 2L

100 22

α = 25

ωo = ωo =

1 LC 1 2 800 × 10 −6

ωo = 25 We see that α = ω0, equivalent to a critically damped circuit. The equation for the current, for this case, is i t = At + B e − αt To find A and B, we apply the circuit’s initial conditions. First Condition: Solve the Current Equation for t = 0 s. Here we go: i t = 0 = A t 0 + B e −αt B=0

0

Solutions

Second Condition: Take the Derivative of the Current Equation and Solve for t = 0 s. We take the derivative of the current equation i t = At + B e − αt d d it = At + B e − αt dt dt Voltage across the inductor is VL = L

di dt

Therefore, di VL = dt L where VL is the voltage across the inductor as the switch is closed, which is known to be equal to the power supply voltage, as explained before. Therefore, d VL 30 d it = = At + B e − αt = dt 2 dt L d At + B e −αt 15 = dt We are before the derivative of a product of functions and we can use the product rule: MATH CONCEPT The derivative of a product of two functions is equal to the derivative of the first function multiplied by the second function added to the first function multiplied by the derivative of the second, or u v = u v + uv

Therefore, d d −αt At + B e −αt + At + B e dt dt d − αt e 15 = Ae − αt + At + B dt

15 =

We already know how to solve the derivative of an exponential; therefore, 15 = Ae − αt + At + B − αe − αt 15 = Ae − αt −α At + B e − αt

445

446

28 RLC Circuits: Part 2

15 = Ae −αt − αAte −αt −αBe − αt By substituting B = 0, we get 15 = Ae −αt − αAte −αt Solving for t = 0 s, 15 = Ae −αt −αA t 0 e − αt 0

0

A = 15 Thus, we can write the final current equation as follows. FINAL RESULT i t = 15te −25t

The current for t = 10 ms is i 10 ms = 15 10 × 10 −3 e −25 10 × 10

−3

FINAL RESULT i 10 ms = 0 1168 A

2

We must find the values of L and C. First, we take the circuit’s second-order differential equation and divide it by 400 to write it in the standard form:

400

d2 i di + 40 + i = 0 2 dt dt

The result is d2 i 1 di 1 + i=0 + dt2 10 dt 400

Solutions

Series RLC circuits are ruled by second-order differential equations in the general form: di2 t R di t 1 i=0 + + 2 L dt LC dt By comparing this equation with the original in the standard form, we can extract the following values: R 1 = L 10 By substituting R = 1 Ω, we get, L = 10 H Another value we can extract is 1 1 = LC 400 Therefore,

C=

400 400 = = 40 F L 10

Knowing R, L, and C, we can find other parameters: α= α=

R 2L

1 2 10

α = 0 05

ωo = ωo =

1 LC 1 10 40

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28 RLC Circuits: Part 2

ωo = 0 05 We see that α = ω0, equivalent to a critically damped circuit. The equation for the current, for this case, is

i t = At + B e − αt To find A and B, we apply the circuit’s initial conditions. First Condition: Solve the Current Equation for t = 0 s. i 0 = 0 = At 0 + B e − αt

0

0 = 0 + B e0

B=0 Second Condition: Take the Derivative of the Current Equation and Solve for t = 0 s. Like we did in the previous example, we take the derivative of the current equation and will get to the following point: di t VL d d = Ate m1 t + Be m2 t = dt L dt dt VL is the voltage across the inductor as the switch is closed, known to be the same as the power supply. Therefore, di t VL 30 d d = Ate m1 t + Be m2 t = = dt L 10 dt dt d d − αt − αt 3 = A − te +B e dt dt In the last example, we already explained how to solve this: 3=A

d d t e −αt + t e −αt dt dt

+B

d − αt e dt

Solutions

3 = A 1 −αte − αt − αBe −αt Solving for t = 0 s, 3 = A − αB By substituting B = 0, A=3 Therefore, the final current for the equation can be written as follows. FINAL RESULT i t = 3te − 0 05t

Current for t = 1 ms is i 1 ms = 3 1 × 10 − 3 e −0 05 1 × 10

FINAL RESULT i 1 ms = 2 99 mA

−3

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29 Transistor Amplifiers The Magic Component

29.1

Introduction

In this chapter, we will examine the usage of transistors as amplifiers. We will show how transistors are able to “amplify”1 signals and will briefly introduce the reader to how they work.

29.2

Transistor as Amplifiers

Nothing is farther from the truth than thinking transistors are magical devices that can amplify the signals they receive, like a magical portal that receives the signal and amplifies it. In fact, transistors are like switches that can manipulate and modulate the voltage coming from the power supply to generate a huge copy of the input signal, creating the illusion of amplification. The output signal is not the input signal amplified but rather the power supply voltage modulated to look like the input signal.

29.3

The Water Storage Tank

Imagine a huge water store tank with a large output pipe. At the bottom of the tank, a tiny water valve is connected electronically to a huge water valve that opens or closes the main pipe. Therefore, if you rotate this tiny valve to one direction, it will make the huge water valve open and let a gigantic amount of water exit from the tank. If you rotate the tiny water valve to the other direction, the huge valve will close. Thus, small rotations of the tiny valve will produce huge rotations of the largest valve. 1 The reader will understand why this word is between quotes later in this book. Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

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29 Transistor Amplifiers

This can be interpreted as an amplification: a small rotation producing a large rotation. This is what roughly happens with transistors: a tiny current flowing into the input will produce a large current flowing across the output.

29.4

Current Gain

If the reader remembers what we have explained previously about transistors, a transistor, which symbol for an NPN type is shown in Figure 29.1, is a threeterminal device, with a collector (C), a base (B), and an emitter (E). In an NPN type transistor, a small current flowing from the base to the emitter (base current) will produce a larger current flowing from the collector to the emitter (collector current). This effect is known as “current gain,” also known as transistor hFE or β. Different transistors offer different values for the current gain, which can vary from 10, 20, and 50 to values like 800. Figure 29.1 NPN transistor.

hFE stands for hybrid parameter, forward transfer, and common emitter.

Therefore, the collector current will be the base current multiplied by hFE, or iC = hFE iB

29.5

Power Supply Rails

All electronic circuits are powered by some kind of voltage source. The voltage source can be a battery or an electrical device of some kind. Suppose this voltage source is a 9 V battery. This battery has a positive and a negative pole. These two poles or voltage limits are known as rails. In an amplifier, the output should never exceed the rails or distortion will appear at the output, and damage can happen to the power supply.

29.6

Amplifying

Suppose we want to amplify a sinusoidal wave like the one shown in Figure 29.2. This wave varies from 1 V to −1 V; therefore, this is a 2 V peak-to-peak voltage, or 2 Vpp. This means the audio signal varies between positive and negative values.

29.7 Quiescent Operating Point

Figure 29.2 2 Vpp audio signal.

Suppose our amplifier is powered by a 9 V battery. If this is the case, the maximum output signal we can collect at the output, in theory, will be a 9 Vpp signal, as shown in Figure 29.3. Figure 29.3 Input signal and maximum output possible.

This happens because the amplifier output is limited by the power supply voltage. Therefore, for an input signal of 2 Vpp, the amplifier will only be able to amplify the signal 4.5 times to get a 9 Vpp output. When the output varies the full range possible, in this case from the battery negative pole or 0 V to the maximum battery voltage, or 9 V, equivalent to the battery positive pole, it is said that the output does a full swing or full rail, going from the minimum to the maximum rail, provided by the power supply. In the real world, due to project and component limitations, amplifiers will never be able to provide a full swing output.

29.7

Quiescent Operating Point

To work as an amplifier, a transistor must be configured properly to operate within a certain range of parameters. This operating point is called “quiescent operating point” or “Q-point.”

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29 Transistor Amplifiers

Adjusting the operating point of a transistor is called biasing. Suppose a transistor amplifier will be used to amplify audio signals. An audio signal has positive and negative values. If our amplifier is powered by 9 V battery and the output cannot go beyond this value, we must bias the transistor in a way that the output sits at 4.5 V, the middle of the rail, when there is no input signal. By biasing the transistor like that, the output will be able to swing from 4.5 to 9 V during the output positive cycle and from 4.5 to 0 V during the negative cycle (see Figure 29.4).

Figure 29.4 Input and output signals.

29.8

Amplifier Classes

Transistors can be used to create amplifiers in very different ways, called “classes.” Every class is more or less energy efficient than another. These classes are named with letters, like A, AB, B, C, D, E, F, G, H, S, and variations of B. The simplest class of them all is class A and, for that reason, will be the theme of this chapter, specifically in a configuration called common emitter, as we will examine later in this chapter. 29.8.1

Class A

Class A amplifiers are probably the cheapest and the simplest to produce. Class A amplifiers are good amplifiers with a relatively low output distortion. The problem with class A is that it consumes too much power. The transistors in a class A configuration are always conducting current, meaning that they are consuming power even if there is no signal at the input. Therefore, they are not the best amplifiers to use in portable devices, powered by batteries. To improve the energy efficiency of class A amplifiers, engineers created other classes that improve the energy problem, but these new classes introduce other problems, like signal distortion.

29.8 Amplifier Classes

Like examined previously in this book, there are two kinds of transistors: NPN and PNP. All explanations in this chapter will be for amplifiers using NPN transistors because all principles can be applied for amplifiers using PNP transistors.

29.8.1.1 Common Emitter

To make a transistor work as an amplifier, it must be properly biased. Biasing a transistor means adjusting its voltages and currents, so it can operate as an amplifier. Common-emitter mode is the most popular way to use a transistor as an amplifier. It is called “common-emitter” amplifier since the emitter is common to both the input and the output circuit. Common-emitter amplifiers have the following characteristics:

•• •• •

Signal enters by the base and exits by the collector. Medium input and output impedances. Medium current and voltage gains. High power gain. Input and output have a phase relationship of 180 .

29.8.1.1.1

Biasing a Transistor

Biasing a transistor to work as a common-emitter amplifier can be done in three basic ways: fixed base biasing, collector feedback biasing, and voltage divider biasing. We will examine these configurations in the next sections. Fixed Base Biasing This method for biasing a common-emitter class A amplifier is seen in Figure 29.5. In Figure 29.5, VCC represents the power supply’s positive pole, and the triangle pointing down, known as the ground point, represents the power supply’s negative pole. RC is the collector resistor and RB is the base resistor. This fixed base biasing configuration uses a resistor connected between VCC and the collector and another one from the same point to the transistor base. The purpose of these resistors is to provide voltage levels and currents for the transistor operation. The emitter is connected to the ground, as shown in Figure 29.5.

Figure 29.5 Fixed base biasing commonemitter amplifier.

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29 Transistor Amplifiers

Transistors are generally represented in diagram by the uppercase letter Q or T.

Figure 29.6 Currents and voltages in a fixed base biasing common-emitter amplifier.

Figure 29.7 Input mesh of a fixed base biasing commonemitter amplifier.

In the common-emitter configuration, the input signal enters by the transistor base, point 1 in Figure 29.5, and exits by the collector (C), or point 2. Therefore, the input signal will be connected between point 1 and ground, and the output will be collected between point 2 and ground. To set the operating point for the fixed base biasing configuration, we start by specifying the currents. Figure 29.6 shows the several currents of this kind of configuration: the collector current (iC), the base current (iB), and the emitter current (iE). Figure 29.6 also shows the voltage across collector– emitter (VCE) and the voltage base–emitter (VBE). Figure 29.6 shows that the base is positive compared with the emitter, meaning that the base must have a greater electrical potential. In fact, to work properly, the voltage at the base must be at least 0.7 V greater than the emitter. If so, current will flow from collector to emitter. The signals of VCE in Figure 29.6 show that the voltage at the collector will be greater than the emitter during normal operation. Currents We start by applying Kirchhoff’s voltage law (KVL) to the circuit’s input mesh shown in Figure 29.7, composed of the base resistor, the power supply, and the base–emitter junction. This mesh is equivalent to the circuit shown in Figure 29.8. If we apply KVL to this mesh, we get

VCC −RB iB − VBE = 0 Therefore, we can find the base current by rearranging the equation RB iB = VCC −VBE Figure 29.8 Equivalent input mesh of a fixed base biasing common-emitter amplifier.

iB =

VCC − VBE RB

29 1

29.8 Amplifier Classes

We know that the collector current is iC = hFE iB

29 2

In the next step, we find the voltage across collector– emitter by applying KVL to the output mesh, shown in Figure 29.9. This output mesh is composed of the collector resistor, the collector–emitter block, and the power supply, as shown in Figure 29.9, and can be redrawn as shown in Figure 29.10. If we apply KVL to this mesh, we get VCC −RC iC − VCE = 0

Figure 29.9 Output mesh of a fixed base biasing commonemitter amplifier.

Therefore, VCE = VCC − RC iC and

iC =

VCC − VCE RC

29 3

A transistor can be one of three states, Figure 29.10 Output mesh namely, cutoff, saturated, or conducting, this equivalent of a fixed base biasing later being equal to the normal operating common-emitter amplifier. mode as an amplifier. If a transistor is saturated, the collector–emitter voltage, VCE, will be 0. Therefore, current will be flowing at maximum intensity, just limited by the collector resistor. In this case, the saturation collector current will be iC sat =

VCC −VCE 0 RC

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29 Transistor Amplifiers

iC sat =

VCC RC

On the other operation extreme, or when the transistor is cutoff or not conducting, VCE will be equal to the power supply voltage, VCC, because there is no current flowing across the collector; therefore,

VCE off = VCC

If we want the transistor to work correctly as an amplifier, we must make sure that it operates in the middle point between saturation and cutoff in what is called the “active region.” We also have to be sure that the input signal does not drive the circuit out of this region. Setting It Up

We will set this project to use the low power general-purpose silicon-based transistor model 2N2222A and to be powered by a 9 V battery. Transistor manufacturers publish datasheets about their products where the electric characteristics of their devices are specified at detail. The datasheet for transistor 2N2222A is available at MIT. You can access their document using our shortcut at http://transistor.katkay.com The datasheet shows the collector–emitter saturation voltage curve, shown in Figure 29.11.

Figure 29.11 Collector–emitter saturation voltage for transistor 2N2222A.

29.8 Amplifier Classes

This graph shows that the collector–emitter voltage keeps its value relatively stable until the collector current reaches about 30 mA. After that, its values increase exponentially. Therefore, we must choose a collector current that is below 30 mA to keep the transistor operating in a stable region. We choose 10 mA for this project. We read the datasheet again and this time we look for the current gain curve (hFE), shown in Figure 29.12.

Figure 29.12 Current gain curve for transistor 2N2222A.

Figure 29.12 shows a value for the hFE parameter equal to 225 for a current of 10 mA at room temperature. Thermal Runaway

Figure 29.12 shows three curves for hFE, one for each different temperature. The higher the temperature, the bigger the hFE value, making it clear that hFE increases when the temperature rises. A transistor will normally heat during normal operation due to the current flowing between collector and emitter. This temperature rise will increase the hFE value. The collector–emitter current is product of the base current and the hFE. If hFE rises, the collector–emitter current will increase and so on, which can lead to the transistor destruction. This effect is called “thermal runaway,” and the best way to prevent this situation is to add a resistor between the emitter and the ground, something that this circuit lacks. The lack of an emitter resistor in this circuit is condition enough not to use this kind of project, but we will show other situations that makes this, in this fixed base biasing method not good as an amplifier.

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29 Transistor Amplifiers

The Design

We have established four values for our amplifier design so far: the power supply voltage, the collector current, the current gain, and the base– emitter voltage: VCC = 9 V iC = 10 mA hFE = 225 VBE = 0 7 V The ideal amplifier would be the one that can provide a full swing output. In this given example, it would be an amplifier that could provide an output varying from 0 to 9 V. But like we have explained before, a few conditions will prevent an amplifier from using the full rail. One of these conditions is the collector–emitter voltage. Reading the graph in Figure 29.11, we see that the transistor 2N2222A has a collector–emitter saturation voltage of, approximately, 0.1 V for 10 mA. As a rule of thumb, we think that it is too dangerous, in terms of transistor stabilization, to have a collector–emitter voltage below 0.7 V, and we recommend always having it bigger than 1 V. To make sure we will have a stable amplifier, we will choose to have this minimum voltage set to 2 V through this chapter. Therefore, the minimum collector–emitter we choose is

VCEMIN = 2 V The emitter is connected to the ground, as we can see in Figure 29.9. Consequently, 2 V is also the minimum collector voltage, or in other words, the collector voltage when the output is at its minimum value. The collector is where the signal output exits, meaning that the output is collected between collector and ground. We have just selected 2 V as the minimum voltage for the collector, meaning that no collector voltage must be below that point. That is the same as saying that the amplifier will only be able to vary from 2 to 9 V. Therefore, the larger output signal may have only the difference between these two values, that is, 7 V peak to peak. Consequently, the output wave will only be able to vary by 3.5 V up and the same amount down. If the maximum allowable voltage for the collector is 9 V, then it must be adjusted to be 3.5 V below that point when at rest, equivalent to 5.5 V = 9 − 3.5, a

29.8 Amplifier Classes

voltage we call “middle point collector voltage,” or “collector at rest voltage,” which can also be found by the following formula: VCmiddle =

VCC − VCMIN + VCMIN 2

or

VCmiddle =

VCC + VCMIN 2

VCC is the power supply voltage. VCMIN is the minimum collector voltage. VCmiddle is the collector voltage at rest.

Therefore, VCmiddle =

9+2 2

VCmiddle = 5 5 V

VC is the collector voltage, measured between the collector and the ground. VCE is the collector–emitter voltage, measured between these two elements.

Now the output can swing 3.5 V up, from 5.5 to 9 V during the positive cycle and 3.5 V down, from 5.5 to 2 V during the negative cycle. Resistors We must calculate, now, the resistors that will make every chosen voltage and current possible. This is what we have chosen so far:

VCC = 9 V iC = 10 mA = 10 × 10 − 3 A VCE = 5 5 V

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29 Transistor Amplifiers

We can start by finding the collector resistor by using Eq. (29.3): RC = RC =

VCC −VCE iC

9−5 5 10 × 10 −3

RC = 350 Ω We can use Eq. (29.2) to find the base current: iC = hFE ib ic iB = hFE iB =

10 × 10 −3 225

iB = 44 44 μA But we also know by Eq. (29.1) that iB =

VCC −VBE RB

and we can use this equation to find the base resistor: RB = RB =

VCC − VBE iB

9 −0 7 44 44 μA

RB = 186 75 kΩ

AC Analysis

the input?

What happens to the circuit when an audio signal is applied to

29.8 Amplifier Classes

First, we must remember that the way we have designed this amplifier was in such a way to establish the currents flowing in the transistor, the base voltage as 0.7 V and the collector voltage at rest, to be at 5.5 V. Suppose we now inject a sinusoidal audio signal at the transistor base, like the one shown in Figure 29.13. This signal can be from a microphone, for example, and varies from +40 to −40 μA. Figure 29.13 Audio signal.

At the beginning, the audio signal sits at 0 μA (see Figure 29.13). Therefore, there is no current injected at the transistor base, and all other currents or voltages in the circuit keep their calculated values. As time passes, the audio signal increases from 0 μA and reaches its peak at +40 μA. We have calculated the base current to be 44.44 μA, but now, there is this new additional current of +40 μA coming from the audio signal, making the base current increase to 84.44 μA. This new base current will make the collector current increase to

The Positive Cycle

iC = hFE iB iC = 225 84 44 × 10 −6

iC = 19 mA In other words, the collector current increased from 10 to 19 mA. If we go back to the output mesh’s KVL equation and solve for VCE, using this new collector current, we get the new collector–emitter voltage: VCE = VCC − RC iC

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29 Transistor Amplifiers VCE = 9 − 350 19 × 10 −3

VCE = 2 35 V In resume, during the positive cycle of the audio signal, it varied from 0 to +40 μA, making the base current increase from 44.44 to 84.44 μA, forcing the collector voltage to vary from 5.5 to 2.35 V. In other words, an increase in the input current produced a decrease in the output voltage, showing that this amplifier is an inverter, meaning that all signals injected at the input will be collected at the output inverted, with a 180 phase offset. The Negative Cycle In this section we will see what happens during the negative cycle of the audio input. The time passes and the audio signal decreases to −40 μA. This new negative current is added to the base current, making it drop from 44.44 to only 4.44 μA. This new current will produce the following collector current:

iC = hFE iB iC = 225 4 44 × 10 −6

iC ≈ 1 mA If we go back again to the output mesh’s KVL equation and solve for VCE using this new current, we get VCE = VCC − RC iC

VCE = 9 − 350 1 × 10 − 3

VCE = 8 65 V

29.8 Amplifier Classes

In resume, the audio signal varied from 0 to −40 μA, making the base current drop from 44.44 to 4.44 μA, forcing the collector voltage to vary from 5.5 to 8.65 V. In other words, a decrease in the input current produced an increase in the output voltage, confirming this amplifier as an inverter. Conclusion This fixed biased transistor amplifier works as an amplifier within certain parameters, but it is rarely used in real life because it has a lot of problems. The first problem is lacking an emitter resistor, which can lead the transistor to self-destruction, for the reasons we have explained. The other problem is that the design is highly dependable on the hFE value. In real life, even the same type of transistor can have different values for hFE. The slightest difference in hFE will result in different voltages and currents across the circuit. To see the dependence of hFE of this kind of circuit, consider the voltage gain of this amplifier. The voltage gain will be the output voltage divided by the input voltage. Looking at Figure 29.6, we can see that the output voltage is VOUT = VCC − RC iC

and that the input voltage is VIN = VCC − RB iB Thus, the voltage gain is VG =

VOUT VCC − RC iC = VIN VCC −RB IB

We have iC and iB in the formula and we know that iC = hFE iB Therefore, the voltage gain is VG =

VCC − RC hFE iB VCC − RB iB

You can see that hFE is in the voltage gain formula and that iB is in both nominator and denominator, making the voltage gain dependent on these values. Even the same kind of transistor hardly has the same hFE. Imagine designing a product to be sold to thousands of customers that behave differently for every person.

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29 Transistor Amplifiers

With this kind of design, voltages and currents do not remain stable during transistor operation and can vary enormously, and even if the transistor does not self-destruct, due to the thermal runaway problem, its temperature will affect and change hFE, increasing the output current and disturbing all other voltages and currents across the circuit. For all these problems, this circuit design is not recommended. In the following section, we will examine other types of transistor biasing and see if we can solve this problem.

Figure 29.14 Collector feedback biasing.

Figure 29.15 Collector feedback biasing (currents).

i = i B + iC

Collector Feedback Biasing Figure 29.14 shows a collector feedback biasing circuit where RC is the collector resistor and RB is the base resistor or the feedback resistor. The base resistor is connected to the collector, instead of being connected to the power supply. This situation creates what is called collector feedback. The word feedback is used because part of the collector current is injected into the base. The collector feedback current ensures that the transistor is always biased in the active region. Like we have seen previously, when the collector current increases, the voltage at the collector drops, because the voltage drop across the collector resistor increases. This reduces the base current that in turn reduces the collector current and so on until it stabilizes at a certain value. We will see in the next paragraphs that this kind of amplifier design is also an inverter amplifier. This feedback resistor is, in fact, creating a negative feedback into the input, because part of the inverted output signal is being injected into the input. There are several benefits of negative feedback in any kind of amplifier like improving system stabilization, reducing distortion and noise, and improving the system bandwidth. However, a reduction in gain is present for all amplifiers using feedback. Currents The first thing is to establish the currents flowing in the circuit, like shown in Figure 29.15. By the diagram in Figure 29.15, we can establish that

29.8 Amplifier Classes

We know that iC = hFE iB Therefore, current across RC will be i = iB + hFE iB

i = hFE + 1 iB

29 4

Consequently, voltage drop across RC is VRC = hFE + 1 iB RC If we apply KVL to the input mesh (Figure 29.16), formed by the collector and the base resistor and the base–emitter junction, we get VCC −VRC − VRB −VBE = 0 which means that the power supply voltage subtracted from the voltage drops across the collector and the base resistors subtracted from the base–emitter voltage is equal to 0. This can be expanded into VCC −iRC −RB iB −VBE = 0 Upon substituting (29.4), VCC − hFE + 1 iB RC − RB iB − VBE = 0 Therefore, the base current will be

Figure 29.16 Input mesh of a collector feedback biasing circuit.

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29 Transistor Amplifiers

iB =

VCC −VBE RB + hFE + 1 RC

29 5

If we apply KVL to the output mesh (Figure 29.17), formed by the collector resistor and the collector– emitter block, we get VCC − VRC − VCE = 0 This can be expanded into

Figure 29.17 Output mesh of a collector feedback biasing circuit.

VCC − RC i− VCE = 0 Upon substituting (29.4), VCC − hFE + 1 iB RC −VCE = 0 Therefore,

iC =

hFE VCC − VBE RB + hFE + 1 RC

29 6

By comparing the base current (29.5) and the collector current (29.6), it is clear that both the collector and the base currents are highly dependable upon hFE. To see the dependence of hFE of this kind of circuit, consider the voltage gain of this amplifier. The voltage gain will be the output voltage divided by the input voltage. Looking at Figure 29.6, we can see that the output voltage is VOUT = VCC −RC iC and that the input voltage is VIN = VCC −RC iC − RB iB Thus, the voltage gain is

29.8 Amplifier Classes

VG =

VOUT VCC − RC iC = VIN VCC − RC iC −RB iB

We have iC and iB in the formula and we know that iC = hFE iB Therefore, the voltage gain is

VG =

VCC − RC hFE iB VCC − RC hFE iB −RB iB

You can see that hFE is in the voltage gain formula and that iB is in both nominator and denominator, making the voltage gain dependent on these values. This design shows the same problem as the fixed base biasing one: currents and voltages will vary wildly if hFE changes. An additional issue with this design is the lack of an emitter resistor to prevent the thermal runaway problem. We must continue our epic journey to find another biasing method that can solve these problems. 29.8.1.1.2

Voltage Divider Biasing

This method of transistor biasing is shown in Figure 29.18, known as voltage divider biasing, and is the most popular method of biasing a transistor. In this kind of biasing, RC is the collector resistor and RB is the base resistor. R1 and R2 are called the bias resistors, configured as a voltage divider that runs a large current across, compared with the base current, and helps to establish a stable base voltage and minimize any variations in the base current. The thermal runaway problem is finally solved by adding an emitter resistor. The emitter resistor RE is, in fact, the sum of two resistors: the emitter resistor RE itself and a resistor we will call “gain resistor,” proper from the transistor, that controls the voltage gain or the amplification factor. We will disregard this resistor in this book because it falls outside the book’s objective. The first thing we must do is to establish the transistor operating point.

Parameters

Figure 29.18 Voltage divider biasing.

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29 Transistor Amplifiers

We have added an emitter resistor to solve the thermal runaway. A current will flow across this resistor and cause a voltage drop. Therefore, the emitter (Figure 29.18) will have a voltage that is not 0, like the previous designs we have examined, because the emitter is not connected to the ground anymore. If our intention is to make an amplifier with the maximum output voltage swing possible, we must choose a value for the emitter voltage that is the lowest we can get, because this voltage will reduce the output swing by the same amount. Remember when we first examined this that the possible output voltage range is VCC subtracted from the collector–emitter voltage. Now with another voltage drop on the output mesh, the maximum output range possible will be VCC subtracted from the collector–emitter voltage and the emitter resistor voltage drop. It is recommended to have the voltage drop across the emitter resistor around 1 V. Like the last time, we chose a collector current of 10 mA, which will result in a value for hFE equal to 225 for transistor 2N2222A. This transistor is silicon based; therefore, its required base–emitter voltage is 0.7 V. We will use a 12 V power supply for this project, instead of a 9 V battery. These are the parameters we have so far: VCC = 12 V iC = 10 mA hFE = 225 VBE = 0 7 V VE = VRE = 1 V Figure 29.19 shows the currents flowing in the voltage divider biasing circuit. To find the current equations, we start by defining the circuit’s input and output meshes, shown in Figure 29.20. By applying KVL to the input mesh, we get

Currents

VCC − VR1 − VR2 = 0 VCC − R1 I1 −R2 I2 = 0

29 7

Doing the same for the output mesh, we obtain

Figure 29.19 Voltage divider biasing (currents).

29.8 Amplifier Classes

Figure 29.20 Voltage divider biasing (input and output meshes).

VCC − VRC − VCE −VRE = 0 VCC −RC IC − VCE − RE IE = 0

29 8

Design Like the previous project, the collector needs a minimum voltage that is recommended not to be less than 2 V. Therefore,

VCEMIN = 2 V If the emitter voltage was chosen to be 1 V and VCEMIN is 2 V, then the collector minimum voltage will be the sum of these two values, or VCMIN = VCEMIN + VRE VCMIN = 2 + 1

VCMIN = 3 V

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29 Transistor Amplifiers

This means that the output voltage will not be able not to vary from 0 to 12 V but rather from 3 to 12 V. Therefore, to have maximum output amplitude, we must set the collector voltage to be at the middle of this range, or in the middle value between 3 and 12 V, or VC = VC =

VCC + VCMIN 2

12 + 3 2

VC = 7 5 V Consequently, the output voltage will be able to vary from 7.5 to 12 V during the positive cycle and from 7.5 to 3 V during the negative cycle, a 4.5 V variation for both sides. We have the collector current and hFE; therefore we can find the base current: iB = iB =

iC hFE

10 × 10 −3 225

iB = 44 4444 μA The emitter current is equal to the sum of collector and base currents, but because the base current is infinitely less than the collector current, we can ignore it and say that, for most purposes, the emitter current is equal to the collector current. However, in this book, we will always consider the base current. Therefore, the emitter current is iE = iC + iB iE = 10 × 10 −3 + 44 4444 × 10 −6

iE = 10 0444 mA Knowing these values, we can find the emitter resistor:

29.8 Amplifier Classes

V R E = R E iE 1 = 10 0444 × 10 − 3 RE

RE ≈ 100 Ω Similarly, the voltage drop across the collector resistor is VRC = RC iC

29 9

The collector voltage is also equal to the sum of the voltage across the collector–emitter and the voltage across the emitter resistor, or VC = VCE + VRE If we substitute this in the output mesh equation in equation (29.8), we get another way to represent the voltage across the collector resistor: VCC − VRC − VCE − VRE = 0 VCC − VRC − VC = 0 VRC = VCC − VC

29 10

which is the power supply voltage subtracted from the collector voltage we have already established before.

Therefore, because Eqs. (29.9) and (29.10) represent the same thing, we can equate them and find the collector resistor: VCC − VC = RC iC 12−7 5 = 10 × 10 − 3 RC

RC = 450 Ω The bias resistors R1 and R2 are configured as a voltage divider. Current i1 flows across R1, and a current i2 flows across R2 (Figure 29.19). Current i1 will be the sum of the base current iB with i2, or

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29 Transistor Amplifiers

i1 = iB + i2 What this means is that the base current will be a fraction of the current flowing across R1. In other words, the base current is drained from the current flowing across R1. We must consider that the currents across R1 and R2 will establish a voltage divider that will establish the base voltage. Therefore, if we want this base voltage to be as invariable as possible, knowing the base current will be drained from the current i1 across R1, we must choose a value for i1 that is much larger than the base current. This will make sure that the base current being drained from i1 will not disturb the voltage set by the divider. As a rule of thumb, we generally choose i1 to be 10 times smaller than the collector current iC, or

i1 = 0 1 iC

We already know the collector current. Therefore, we can calculate i1: i1 = 0 1 10 mA

i1 = 1 mA We know that the base voltage will be equal to the sum of the emitter voltage and the base–emitter voltage, which, for silicon-based transistors, is approximately equal to 0.7 V. We also know the emitter voltage; therefore, we can find the base voltage: VB = VBE + VE VB = 0 7 + 1

VB = 1 7 V In Figure 29.19, we see that the base is connected between resistors R1 and R2, as shown in Figure 29.21.

Figure 29.21 Voltage divider biasing – The base.

29.8 Amplifier Classes

Resistor R2 is connected between the base and ground. The voltage across this resistor is equal to the product of its resistance and current, which is also the base voltage in relation to ground, or V R 2 = R 2 i2 = V B We have two equations for VA. Consequently we can equate them: R 2 I2 = V B We already know that i1 = 1 mA iB = 44 4444 μA and i1 = i2 + iB Therefore, i2 = i1 −iB i2 = 1 mA−44 4444 μA i2 = 10 −3 −44 4444 × 10 −6 i2 = 1 mA−44 4444 μA

i2 = 955 5555 μA We can now find R2: VR2 = R2 i2 VR VB R2 = 2 = i2 i2 VB R2 = i2 Substituting VR2 = VB = 1 7 V i2 = 955 5555 μA R2 =

17 955 5555 × 10 −6

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29 Transistor Amplifiers

R2 = 1779 06 Ω To find R1, we go back to the input mesh equation: VCC − VR1 − VR2 = 0 We already know that V R2 = V B = 1 7 V VCC = 12 V Therefore, VR1 = VCC −VR2 VR1 = 12− 1 7

VR1 = 10 3 V We already know i1, and therefore we can find R1: VR1 = R1 i1 VR R1 = 1 i1 R1 =

10 3 1 × 10 − 3

R1 = 10 3 kΩ In resume, these are the values we have designed so far: R1 = 10 3 kΩ R2 = 1779 06 Ω RC = 450 Ω RE = 100 Ω However, when choosing components like resistors, capacitors, inductors, and so on, we must rely on values that are available for sale on the market and some of the calculated values are not.

Exercises

Checking Appendix G, we choose the final values for 20% tolerance resistors: R1 = 10 kΩ R2 = 1 5 kΩ RC = 470 Ω RE = 100 Ω These new values will change voltages and currents across the circuit. By the end of this chapter, at the exercises section, we will examine the effect of such changes. 29.8.1.1.3

Conclusion

One thing that should be observed about the voltage divider method is that the base current and the base voltage depend basically on the currents circulating through R1 and R2 and the voltages established by these resistors are independent from hFE. To see the dependence of hFE of this kind of circuit, consider the voltage gain of this amplifier. The voltage gain will be the output voltage divided by the input voltage. Looking at Figure 29.19, we can see that the output voltage is VOUT = VCC − RC iC and the input voltage is VIN = VCC ×

R2 R1 + R2

Therefore, the voltage gain is VG =

VOUT VCC − RC iC × R1 + R2 = VIN VCC R2

We do not see hFE on the voltage gain formula because there is no dependence on the base current too, showing how the gain is independent from hFE.

Exercises 1

Consider the voltage divider biasing common-emitter class A amplifier shown in Figure 29.22. The circuit uses the NPN silicon-based low power general-purpose transistor type 2N2222A.

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29 Transistor Amplifiers

Figure 29.22 Voltage divider biasing.

Figure 29.23 shows the hFE curve for this transistor.

Figure 29.23 hFE DC current gain (transistor 2N2222A).

The power supply voltage and the collector current should be VCC = 18 V iC = 8 mA Design this amplifier for maximum output swing possible.

Solutions

Solutions 1

The first thing we do is to define a voltage for the emitter. This will be the voltage across the emitter resistor and as a rule of thumb should be around 1 V to keep the amplifier stable and prevent the thermal runaway. The current collector was chosen to be 8 mA. This value gives us a value of hFE equal to 225 for this 2N2222A transistor. This is a silicon transistor; therefore, the base–emitter voltage is approximately 0.7 V. These are the values we have so far: VCC = 18 V iC = 8 mA hFE = 225 VBE = 0 7 V The base current can be calculated: iB = iB =

iC hFE

8 × 10 − 3 225

iB = 35 5555 μA We have chosen the emitter voltage to be 1 V. As a rule of thumb, we know that the collector–emitter voltage should be at least 2 V to keep the transistor out of the saturation zone: VCEMIN = 2 V Therefore, the minimum value for the collector voltage during the negative cycle of the output will be VCMIN = VCEMIN + VE VCMIN = 2 + 1

VCMIN = 3 V

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29 Transistor Amplifiers

This means that the collector will be able to swing from 3 to 18 V and consequently will have an amplitude of 15 V peak to peak, equivalent to the subtraction of 18 and 3 V. Therefore, for maximum swing, we must position the collector voltage at a middle point between 3 and 18 V, or VC = VC =

VCC + VCMIN 2

18 + 3 2

VC = 10 5 V The collector sitting at 10.5 V will be able to swing from 10.5 to 18 V during the positive cycle and from 10.5 to 3 V during the negative cycle of the output signal. The emitter current is the sum of the collector current and base current, or i E = iC + iB iE = 8 × 10 −3 + 35 5555 × 10 −6

iE = 8 0355 mA Now we can find the emitter resistor: VRE = iE RE 1 = 8 0355 × 10 −3 RE

RE = 124 4447 Ω Similarly, the voltage drop across the collector resistor is VRC = iC RC and is also VRC = VCC − VC

Solutions

Therefore, we can equate these two equations and find the collector resistor: VCC − VC = RC iC 18−10 5 = RC 8 × 10 − 3

RC = 937 5 Ω R1 and R2 are a voltage divider. As a rule of thumb, the current across R1 should be 10 times less than the collector current. We can use this to get the current across R1: i1 = i1 =

iC 10

8 × 10 −3 10

i1 = 800 μA We have defined the voltage across the emitter resistor to be 1 V. We know that the voltage across base–emitter, for silicon-based transistors, is around 0.7 V. Therefore, we can find the base voltage: VB = VBE + VE VB = 0 7 + 1

VB = 1 7 V R2 is connected between the base and ground. Therefore, the voltage across this resistor is the same as the base voltage, or V R2 = V B Therefore, R 2 i2 = V B VB R2 = i2

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29 Transistor Amplifiers

But we know that i 2 = i1 − iB Therefore, we can find R2: R2 =

VB i1 − iB

R2 =

17 800 × 10 −6 − 35 5555 × 10 −6

R2 ≈ 2224 Ω If R1and R2 are a voltage divider and we know the voltage between them and the current flowing, we can use this to find R1: R1 = R1 =

VCC −VB i1

18−1 7 800 × 10 −6

R1 = 20375 Ω We have finished the project. These are the values we have so far: R1 = 20375 Ω R2 = 2224 Ω RC = 937 5 Ω RE = 124 4447 Ω However, most of these values do not exist in the market. We must replace them with the available values listed on Appendix G. The next list contains the values we have chosen for 20% tolerance resistors: R1 = 18 kΩ or 22 kΩ R2 = 2 2 kΩ RC = 1 kΩ RE = 120 Ω

Solutions

The calculated value for R1 = 20375 Ω is approximately equidistant of the available values of 18 and 22 kΩ. For this reason, we are not exactly sure which one we must choose. To know which one is the best one for our purposes, let us test both options using a software simulator.2 The results provided by the simulator were the following: Choosing R1 equal to 18 kΩ will increase the current across R1 from 800 μA to about 900 μA and an increase on the current across R2 from 764 μA to around 852 μA. These two changes modify the base voltage from 1.69 to 1.87 V. Collector voltage drops from 9.81 to 7.52 V and the emitter voltage rises from 1.08 to 1.3 V. In technical terms, the output swing is now limited to vary from 3.3 to 18 V. This is not the middle point between 3.3 and 18 V. If the collector voltage is now sitting at 7.52 V, it means that it can vary 10.48 V up, from 7.52 to 18 V, and just 4.22 V down, from 7.52 to 3.3 V. Because the output amplification must always be equally up and down, we must limit the amplification to swing 4.22 V up and down. So, the maximum output voltage we may have without clipping is a wave with 8.44 V peak to peak. However, if we choose R1 to be 22 kΩ, the simulator tells us that the current across R1 will drop from 800 to 746 μA and the current across R2 will drop from 764 to 714 μA. These changes will modify the base voltage from 1.7 to 1.57 V. The collector voltage will rise from 9.81 to 10.01 V, and the emitter voltage will drop from 1 V to 961 mV. The reader can already realize the changes now are less dramatic, compared with what we have designed. The collector voltage sitting at 10.01 V provides output variations of 8 V up to 18 V and 7.049 V down to 2.961 V. For the same reasons as before, the output must be limited to 7.049 V up and down, making the total output’s peak-to-peak amplitude equal to 14.098 V, almost twice the previous case. Therefore, choosing R1 equal to 22 kΩ makes a much better amplifier, closer to what we have designed.

2 www.systemvision.com, for example, or any other the reader prefers.

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30 Operational Amplifiers A Brief Introduction

30.1

Introduction

In this chapter, we will do a brief explanation about operational amplifiers.

30.2

Operational Amplifiers

An operational amplifier or an op-amp is a high-gain voltage amplifier with a differential input and, generally, a single output. In this configuration, an op-amp produces an output voltage that is commonly hundreds of thousands of times larger than the difference voltage between its input terminals. Figure 30.1 shows the symbol for the operational amplifier. Figure 30.1 Operational amplifier.

V+ is the non-inverting input. V− is the inverting input. VS+ is to be connected to the positive power rail from the power supply. VS− is to be connected to the negative power rail from the power supply. VOUT is the output.

Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

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30 Operational Amplifiers

30.3

How Op-Amp Works

An op-amp is basically a voltage amplifier to be used with external feedback components like resistors and capacitors between its inputs and output terminals. These components will determine the operation to be performed by the amplifier, for example, sum, subtraction, integration, differentiation, etc., which is the reason why it is called operational amplifier. Op-amps have two inputs, a non-inverting input and an inverting input. These inputs are called differential inputs. Internally they are as shown in Figure 30.2. Figure 30.2 Operational amplifier (internal circuit).

Signals applied to the non-inverting input will appear at the output amplified but with the same phase as the input, like shown in Figure 30.3. Figure 30.3 Non-inverting input.

However, if the same signal is applied to the inverting input, it will appear 180 inverted in phase at the output, as shown in Figure 30.4.

30.3 How Op-Amp Works

Figure 30.4 Inverting input.

As mentioned earlier, op-amps have differential inputs, meaning that if the same signal is applied at both of its inputs, the output will be the difference, or the subtraction of both signals, as shown in Figure 30.5. Figure 30.5 Differential inputs.

In Figure 30.5, the line at the output can be interpreted as 0 or as no change, meaning that the output will be constant if both inputs are the same. This is known as “common mode rejection ratio.” VS+ and VS− terminals should be connected to the positive and negative power rails. Op-amps generally like to operate with symmetrical power supplies, meaning that the power supply should provide a positive voltage, a negative voltage, and a ground reference. If the reader remembers the mountain analogy, negative and positive voltages are relative and, therefore, can be simulated by, for example, establishing a voltage divider using two resistors, like shown in Figure 30.6. Figure 30.5 shows two resistors in series connected to a 9-V battery to create a voltage divider and simulate a symmetrical power supply with +4.5 V, −4.5 V, and ground (GND) in a configuration known as virtual ground. This method is very limited, and any current gain above certain levels will make the ground level to change, unbalancing the voltages.

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30 Operational Amplifiers

Figure 30.6 Virtual ground.

30.4

•• •• • •

Op-Amp Characteristics

The voltage gain is very high, something that can go up to 10000 or more. Very high input impedance, generally near infinity, resulting in basically no current flowing into either of its two inputs and zero offset between the same inputs. Very low output impedance, generally near 0. The output of an op-amp will be the difference between the voltage signals applied to its inputs multiplied by a certain gain. The gain of an op-amp will be infinite if there is no element connecting the input and the output. In this case, the gain is known as “open-loop gain.” If an element, like a resistor, is connected between the input and the output, the total gain will be reduced and therefore can be adjusted to a predetermined level.

30.5

Typical Configurations

An op-amp can be basically configured as an inverting op-amp and a noninverting op-amp. 30.5.1

Inverting Op-Amp

Figure 30.7 shows a typical inverting op-amp circuit.

30.5 Typical Configurations

Figure 30.7 Inverting op-amp.

To find the output voltage and consequently the gain, let us establish some points and currents, as shown in Figure 30.8. Figure 30.8 Inverting amplifier (currents).

By using Kirchhoff’s current law, we know that

iIN + iF = iX iIN is the input current. iF is the feedback current. iX is the inverting input current.

and that the current flowing across an op-amp input is extremely low, is negligible, and can be considered 0. Therefore, iIN + iF = 0 iIN = − iF

30 1

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30 Operational Amplifiers

We get the feedback current by applying nodal analysis to the output node:

iF =

VOUT − VX RF

The same nodal analysis can be applied to node X to get iIN: iIN =

VIN − VX RIN

We can substitute both equations in (30.1) and obtain VIN −VX VOUT − VX =− RIN RF

30 2

We know that the difference between the op-amps inputs is 0, and we notice that the non-inverting input is connected to ground. Therefore, the voltage across the non-inverting input is 0, and consequently the voltage of the inverting input will also be 0, or VX = 0 Substituting this in (30.2), VIN − 0 VOUT −0 =− RIN RF VIN VOUT =− RIN RF We get the final output voltage. INVERTING OP-AMP OUTPUT VOLTAGE VOUT = −

RF VIN RIN

VIN is the input voltage. VOUT is the output voltage. RF is the feedback resistor. RIN is the input resistor.

Consequently, the gain is

30.5 Typical Configurations

INVERTING OP-AMP VOLTAGE GAIN AV =

VOUT RF =− VIN RIN

AV is the voltage gain. VIN is the input voltage. VOUT is the output voltage. RF is the feedback resistor. RIN is the input resistor.

30.5.2

Non-inverting Op-Amp

Figure 30.9 shows a typical non-inverting op-amp circuit. Figure 30.9 Non-inverting op-amp.

This kind of amplifier will amplify the signal and keep the signal with the same phase as the input. To find the output voltage and consequently the gain, let us establish some points and currents, as shown in Figure 30.10. Figure 30.10 Non-inverting amplifier (currents).

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30 Operational Amplifiers

By applying Kirchhoff’s current law in node Y, we get iA = iF + iY iIN is the input current. iF is the feedback current. iX is the inverting input current.

Like in the last example, we know that the current flowing across an op-amp input is extremely low, is negligible, and can be considered 0. Therefore, i A = iF + 0

iA = i F

30 3

We get the feedback current by applying nodal analysis to the output node:

iF =

VOUT − VY RF

The same nodal analysis can be applied to node Y to get iA:

iA =

VY RA

We can substitute both equations in (30.3): VOUT − VY VY = RF RA VOUT − VY RA VY = RF VY RF = VOUT − VY RA VY RF = VOUT RA − VY RA VOUT RA = VY RF + VY RA VY RF + VY RA VOUT = RA

30.5 Typical Configurations

VOUT = VY

RF + RA RA

VOUT = VY

RF +1 RA

But we know that for op-amps the input voltages are the same; therefore, VY = VIN Consequently, we get the final formula for the output voltage, INVERTING OP-AMP OUTPUT VOLTAGE VOUT = VIN

RF +1 RA

AV is the voltage gain. VIN is the input. VOUT is the output. RF is the feedback resistor. RA is the voltage divider resistor.

and for the gain, INVERTING OP-AMP GAIN AV =

VOUT RF =1+ VIN RA

AV is the voltage gain. VIN is the input. VOUT is the output. RF is the feedback resistor. RA is the voltage divider resistor.

30.5.3

Voltage Follower

Figure 30.11 shows an amplifier where the feedback resistor, RF, was replaced with a wire. This kind of configuration is called “voltage follower” or “buffer,” also known as “unity-gain amplifier” or “isolation amplifier,” because this amplifier has a gain of 1, as follows:

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30 Operational Amplifiers

Figure 30.11 Voltage follower.

AV =

VOUT RF =1+ VIN RA

But RF = 0 therefore, AV = 1 +

0 RA

AV = 1 + 0

VOLTAGE FOLLOWER OP-AMP GAIN AV = 1

A voltage follower does not provide any amplification to the signal. The reason it is called a voltage follower is because the output voltage directly follows the input voltage, meaning the output voltage is the same as the input voltage. A voltage follower acts as an isolation, a buffer, providing no amplification or attenuation to the signal and can be used, for example, as an interface to interconnect two circuits, providing isolation between the two. One example of the advantage of having isolation can be explained using the voltage divider circuit we have mentioned in Figure 30.6. As explained earlier, this virtual ground works by the principle that the 9 V provided by the battery will be split into +4.5 V, −4.5 V, and ground. This will only work and keep the symmetry if the ground value remains stable at 0 V. The big problem of this circuit is that the negative and the positive part of the power supply must supply circuits that drain the same current from both sides. If this is not true, the virtual ground will move from zero and the power supply will not be symmetrical anymore.

30.5 Typical Configurations

Let us see what happens if we connect a load of 10 kΩ to the negative part, as shown in Figure 30.12.

Figure 30.12 Virtual ground plus load.

Resistors R2 and RL are now in parallel. According to the first Ohm law, they can be replaced by an equivalent: 1 1 1 = + REQ R2 RL 1 1 1 = + REQ 100000 10000

REQ = 9090 90 = 90 9090 kΩ The voltage divider will now be equal to what is shown in Figure 30.13. Figure 30.13 Voltage divider as virtual ground.

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The voltage divider that was composed of two 100 kΩ resistors is now composed of a resistor R1 of 100 kΩ and another resistor, REQ, of approximately 9 kΩ. This means that the voltage between ground and V− that was −4.5 V before is now just −0.75 V or six times less. If the voltage dropped in the negative part, it raised on the positive side. Now the voltage between ground and V+ that was +4.5 V is now +8.25 V. The conclusion is that the ground voltage is not in the middle between V− and V+, anymore. Thus, a voltage divider using resistors is not a good solution to build a symmetrical power supply from a nonsymmetrical voltage source. We need something that can isolate the load from the divider. Any ideas how we can accomplish this? The solution to our problem is the one shown in Figure 30.14.

Figure 30.14 Voltage divider improved with a voltage follower.

Now, the voltage follower input is connected between the load RL and the voltage divider, isolating one from the other. The current drained by the load will not affect the voltage divider that much and will keep its value stable. This solution is not completely perfect but is better than having the voltage divider alone. This solution works because the op-amp offers a very high input impedance and, for that reason, will drain very little current from the voltage divider. With a very high input impedance, something around 100 MΩ or more, it is now that impedance in parallel with R2 that gives an equivalent resistor still equal to R2, keeping the relationship between R1 and R2 the same and consequently the voltage balance stable. The following are simple examples of circuits using op-amps.

30.5 Typical Configurations

30.5.4

Non-inverting Summing Amplifier

Figure 30.15 shows a non-inverting summing amplifier that can be used to add multiple voltages. Figure 30.15 Non-inverting summing amplifier.

To find the output voltage, let us establish some points and currents, as shown in Figure 30.16. Figure 30.16 Non-inverting summing amplifier (currents).

We have already calculated previously in this chapter that the output voltage for the non-inverting amplifier is

VOUT = 1 +

RF VIN RA

or in this given example,

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30 Operational Amplifiers

VOUT = 1 +

RF VX RA

30 4

We know that both op-amps have the same voltage, or VX ≈ VY Therefore, the following formula is also true:

VOUT = 1 +

RF VY RA

We also know that the sum of the input currents is equal to the current flowing into the non-inverting input, or i1 + i 2 + i 3 + i 4 +

+ iN = iX

However, the current flowing across an op-amp input is extremely low and negligible, and we can consider it to be 0. Therefore, i A = iF + 0

i1 + i 2 + i 3 + i 4 +

+ iN = 0

30 5

We get each current by applying nodal analysis to node X, for each input. Therefore, VX −V1 R1 VX −V2 i2 = R2 VX −V3 i3 = R3 VX −V4 i4 = R4 i1 =

and so on for the multiple inputs.

30.5 Typical Configurations

By substituting these equations into (30.5), we obtain VX −V1 VX −V2 VX − V3 VX − V4 + + + + R1 R2 R3 R4

+

VX − VN =0 RN

We can expand this equation into VX V1 VX V2 VX V3 VX V4 − + − + − + − + R1 R1 R2 R2 R3 R3 R4 R4

+

VX VN − =0 RN RN

Isolating all VX on one side, VX VX VX VX + + + + R1 R2 R3 R4

VX

+

VX V 1 V2 V 3 V4 = + + + + RN R1 R2 R3 R4

1 1 1 1 + + + + R1 R2 R3 R4

+

+

VN RN

1 V 1 V2 V 3 V4 = + + + + RN R1 R2 R3 R4

+

VN RN 30 6

We notice that the block 1 1 1 1 + + + + R1 R2 R3 R4

+

1 RN

is equal to the equivalent resistance of resistors in parallel and we can, consequently, replace the whole block with that resistance, or 1 1 1 1 + + + + R1 R2 R3 R4

+

1 1 = RN REQ

Therefore, by substituting this into (30.6), we get VX

1 V 1 V2 V 3 V 4 + + + + = REQ R1 R2 R3 R4

+

VN RN

V 1 V 2 V 3 V4 + + + + R1 R2 R3 R4

+

VN RN

VX = REQ

30 7

Upon substituting (30.7) in (30.4), we get the final equation for the output voltage.

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NON-INVERTING SUMMING AMPLIFIER – OUTPUT VOLTAGE VOUT = 1 +

RF V 1 V2 V 3 V 4 REQ + + + + RA R1 R2 R3 R4

+

VN RN

30 8

VOUT is the output voltage. RF is the feedback resistor. RA is the grounding resistor. REQ is the equivalent resistance to all input resistors in parallel. V1 , V2 ,… are the individual input voltages. R1 , R2 ,… are the individual input resistors.

30.5.4.1 Summing Two Inputs

If we are summing just two signals and the input resistors are the same, a very typical case in the real world, the output voltage will be VOUT = 1 +

RF V 1 V2 REQ + RA R R

VOUT = 1 +

RF V1 + V2 REQ RA R

30 9

We know that the equivalent resistor of two resistors in parallel is REQ =

R1 × R2 R1 + R2

For two equal resistors, we have R×R R+R R2 REQ = 2R

REQ =

REQ =

R 2

By substituting into (30.9), we get the final formula for the output voltage for two inputs, VOUT = 1 +

RF RA

R 2

V1 + V2 R

30.5 Typical Configurations

VOUT = 1 +

RF RA

R 2

V1 + V2 R

NON-INVERTING SUMMING AMPLIFIER – OUTPUT VOLTAGE FOR TWO INPUTS VOUT = 1 +

RF RA

V1 + V2 2

VOUT is the output voltage. RF is the feedback resistor. RA is the grounding resistor. V1 , V2 , … are the individual input voltages.

30.5.4.2 Summing Three Inputs

If we have three inputs, the result is slightly different.

VOUT = 1 +

RF V1 V 2 V 3 + + REQ RA R R R

30 10

We know that the equivalent resistor of three resistors in parallel is 1 1 1 1 = + + REQ R1 R2 R3 If we solve that for equal resistors, we have 1 1 1 1 = + + REQ R R R 1 3 = REQ R

REQ =

R 3

By substituting into (30.10), we get the final formula for the output voltage for three inputs,

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30 Operational Amplifiers

VOUT = 1 +

RF RA

R 3

V1 + V2 + V3 R

VOUT = 1 +

RF RA

R 3

V1 + V 2 + V 3 R

NON-INVERTING SUMMING AMPLIFIER – OUTPUT VOLTAGE FOR THREEE INPUTS VOUT = 1 +

RF RA

V1 + V2 + V3 3

V1 , V2 ,… are the individual input voltages. VOUT is the output voltage. RF is the feedback resistor. R is the input resistor.

30.5.5

Inverting Summing Amplifier

This case is like the previous one, where we want to sum several input signals. However, this is an inverter amplifier, where the output is 180 out of phase, compared to the input. To sum more than two voltages, the inverting summing amplifier circuit, as shown in Figure 30.17, is the best one to choose. Figure 30.17 Inverting summing amplifier.

To find the output voltage, let us establish some points and currents, as shown in Figure 30.18.

30.5 Typical Configurations

Figure 30.18 Inverting summing amplifier (currents).

We have already calculated previously in this chapter that the output voltage for the inverting amplifier is

VOUT =

RF VIN RA

or in this given example,

VOUT =

RF VX RA

30 11

We know that both inputs have the same voltage. Because the non-inverting input is connected to ground, we have VX ≈ VY = 0

30 12

We know that the sum of the input currents is equal to the current flowing into the inverting input, or i1 + i2 + i3 + i4 +

+ iN + iF = iX

However, the current flowing across an op-amp input is extremely low and negligible, and we can consider it to be 0.

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30 Operational Amplifiers

Therefore, i1 + i2 + i3 + i4 +

+ iN + iF = 0

i 1 + i2 + i3 + i4 +

+ iN = − i F

iF = − i 1 + i2 + i3 + i4 +

+ iN

30 13

We get each current by applying nodal analysis to node X, for each input. Therefore, V1 − VX R1 V2 − VX i2 = R2 V3 − VX i3 = R3 V4 − VX i4 = R4 VOUT − VX iF = RF i1 =

and so on for the multiple inputs. By substituting (30.12) into these equations, we get V1 R1 V2 i2 = R2 V3 i3 = R3 V4 i4 = R4 VOUT iF = RF

i1 =

30.5 Typical Configurations

Therefore, Equation (30.13) is now VOUT V 1 V 2 V 3 V4 =− + + + + RF R1 R2 R3 R4

+

VN RN

and we get the final equation for the output voltage. INVERTING SUMMING AMPLIFIER – OUTPUT VOLTAGE VOUT = − RF

V 1 V2 V 3 V4 + + + + R1 R 2 R3 R4

+

VN RN

V1 , V2 , … are the individual input voltages. VOUT is the output voltage. RF is the feedback resistor. R1 , R2 , … are the individual input resistors.

30.5.6

Integrator

Like explained at the beginning of this chapter, operational amplifiers have this name because they can perform mathematical operations. Performing the integral of the input signal is just one of these operations. The output of this circuit is the integral of the input (see Figure 30.19). Figure 30.19 Op-amp integrator.

To find the output voltage equation, we start by specifying some currents and points, like shown in Figure 30.20. By applying Kirchhoff’s current law to node X, we get i R + iC = iX However, we know that the current flowing across any of the op-amp inputs is negligible and can be considered to be 0.

505

506

30 Operational Amplifiers

Figure 30.20 Op-amp integrator (currents).

Therefore, i R + iC = 0 iR = −iC

30 14

By applying nodal analysis to node X, we find the current across R: iR =

VIN − VX R

We know that both op-amp inputs have the same voltage, which in this case is 0, because the non-inverting input is connected to ground. Thus, VX = 0 and consequently,

iR =

VIN R

For the capacitor case, we know that the current flowing across this element is

i=C

dV dt

The voltage across the capacitor is VOUT − VX Therefore, the current is iC = C

d VOUT − VX dt

30.5 Typical Configurations

By substituting VX = 0 we get iC = C

dVOUT dt

We can substitute iRand iF in (30.14): iR = − i C VIN dVOUT = −C R dt or, dVOUT VIN =− dt RC If we integrate both sides, we get the formula for the voltage output. INTEGRATOR – OUTPUT VOLTAGE VOUT = −

1 VIN dt RC

VIN is the input voltage. VOUT is the output voltage. R is the input resistor. C is the feedback capacitor.

30.5.7

Differentiator

The differentiator circuit, able to differentiate the input signal, is shown in Figure 30.21.

Figure 30.21 Differentiator.

507

508

30 Operational Amplifiers

To get the output voltage equation, we do the same thing we did in the previous example to get iC = − i R where

iC = C

iR =

dVIN dt

VOUT R

Therefore, C

dVIN VOUT =− dt R

We get the output voltage equation as follows. DIFFERENTIATOR – OUTPUT VOLTAGE VOUT = − RC

dVIN dt

VIN is the input voltage. VOUT is the output voltage. R is the feedback resistor. C is the input capacitor.

509

31 Instrumentation and Bench A Brief Introduction

31.1

Introduction

In this chapter, we introduce the reader to instrumentation, a bench equipment, essential to those willing to project and build electronic circuits.

31.2

Multimeter

A multimeter is a test tool used to measure multiple electrical values, principally voltage (Volts), current (Amperes), and resistance (Ohms). Modern multimeters generally group a lot more functions, and some models can measure frequency (Hertz), capacitance (Farads), and inductance (Henries) and perform tests on diodes and transistors. Modern multimeters are digital devices, like the one shown in Figure 31.1, but, in the past, they were analog needle-based mechanical devices.

Figure 31.1 Digital multimeter.

Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

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31 Instrumentation and Bench

Multimeters are formed by three basic parts:

•• •

Display (A), where measurement readouts can be viewed. Dial or rotary switch (B) for selecting primary measurement values (voltage, current, frequency, etc.). Input jacks (C), where the test leads are inserted.

31.2.1

True RMS Multimeter

Measuring AC currents and voltages can be a problem if a multimeter is not prepared to do that. An average responding multimeter uses average mathematical formulas to measure pure sinusoidal waves with a certain accuracy. If the wave being measured is not sinusoidal, like a square or triangular wave, the measurement accuracy decreases immensely. In many cases, a regular multimeter will offer an accuracy that can be 40% off the real value. The usage of non-sinusoidal waves increased greatly in recent years in fields like robotics, computers, control of variable speed motors, and more. Therefore, it is essential to have a multimeter that can measure such wave forms with precision, like true root mean square (RMS) multimeters can do.

31.3

Voltmeter

Voltmeters are tools that measure the difference in potential (voltage) between two points. Voltmeters are part of multimeters. To use the voltmeter part of a multimeter, its rotary switch must be rotated to the correspondent position, like illustrated with the V letter in Figure 31.1.

Some multimeters may have two or more positions for voltages: VDC and VAC for direct and alternating voltages.

To measure voltage, the voltmeter test leads must be connected in parallel with the circuit being measured. Figure 31.2 shows a multimeter measuring the voltage of a battery.

31.4 Ammeter

Figure 31.2 Voltmeter measuring a battery.

31.4

Ammeter

Ammeters are tools that measure the current flowing across two points in a circuit. Ammeters are also part of multimeters. To use the ammeter part of a multimeter, its rotary switch must be rotated to the correspondent position, like illustrated with the A letter in Figure 31.3.

Figure 31.3 Ammeter measuring current.

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512

31 Instrumentation and Bench

Some multimeters may have two or more positions for current: DC and AC, for direct and alternating currents.

To measure current, the voltmeter’s test leads must be connected in series with the circuit being measured. Figure 31.3 shows a multimeter measuring the current flowing in a circuit with a battery and a light bulb.

31.5

Ohmmeter

Ohmmeters are tools that measure the electrical resistance between two points in a circuit. Ohmmeters are also part of multimeters. To use the ohmmeter part of a multimeter, its rotary switch must be rotated to the correspondent position, like illustrated with the Ω letter in Figure 31.4.

Figure 31.4 Ohmmeter measuring a resistor.

Some multimeters may have two or more positions for resistance, one for every range of resistances.

To measure resistance, the multimeter test leads must be connected between two points in a circuit, and the circuit must be turned off, without any power supply or stored charge.

31.7 Breadboards

A multimeter can be easily damaged if connected to a circuit that is being powered by a power supply or has charge stored in capacitors.

Figure 31.4 shows a multimeter measuring the resistance of a resistor.

31.6

Oscilloscope

Oscilloscopes are an important tool in any electrical engineer lab, because they allow you to see electric signals as they vary over time. Figure 31.5 shows an illustration representing the main parts of an oscilloscope: Figure 31.5 Oscilloscope.

•• •

Display (A), where measurement readouts and the waveforms can be viewed. Buttons where the amplitude or width of the waveform can be adjusted for every channel. Input jacks (C), where the test leads for every channel are inserted.

Oscilloscopes are a kind of voltmeter where you can see the waveform being measured. However, oscilloscopes have channels that allow multiple waveforms to be viewed and measured at the same time in the same display. Oscilloscopes are used in the same way you would use a voltmeter.

31.7

Breadboards

Electrical circuits in their physical form are usually formed by components soldered to a fiberglass board containing faces covered with copper tracks, which interconnect the several points of a circuit. A breadboard, on the other hand, is a solderless reusable device for temporary prototype with electronics and test circuit designs. This makes it easy to use for creating temporary prototypes and experimenting with circuit design.

513

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31 Instrumentation and Bench

Electronic components can be interconnected by inserting their leads or terminals into the holes of a breadboard. The board has strips of metal underneath the board, connecting rows of pins. Figure 31.6 shows an illustration representing the parts of a typical breadboard:

Figure 31.6 Breadboard.

••

Component area (A), where components are inserted. Power and other buses (B), where ground, power supply, and other elements are established.

The rows and columns inside areas A and B are internally connected as shown in Figure 31.7.

Figure 31.7 Breadboard (internal connections).

31.8 Wire Diameter

Examining Figure 31.7, we see that all holes in row A are electrically connected, from column 0 through 29. Consequently, a component inserted at hole A0 and another one at hole A29 will be electrically connected. The same is true for rows B, M, and N. However, every column from 0 to 20 is electrically connected from C to G and from H to L. In column 0, for example, holes C0, D0, E0, F0, and G0 are connected to each other. Following that, holes H0, I0, J0, K0, and L0, are also connected to each other. However, these two groups of five holes are not connected, being insulated from each other. If a component is inserted at C0 and another one at G0, they will be electrically connected. However, if we insert a component at G0 and another one at H0, they will not be electrically connected. The same is true for all holes in lines from C to L in Figure 31.7. Figure 31.8 shows a picture of a typical breadboard in use.

Figure 31.8 Breadboard in use.

31.8

Wire Diameter

Most wires are, in general, cylindrical tubes covered with insulating plastic. Due to electrical resistance, current flowing through a wire makes the wire heat. According to the second Ohm’s law, the resistance in a wire is inversely proportional to the wire’s cross-sectional area, or R=

ρL A

515

516

31 Instrumentation and Bench

For that reason, the smaller the wire’s cross section, the lesser the current a wire can handle without overheating. Appendix F shows the American standard for wire gauges or American Wire Gauges (AWG) and the international millimetric standard. In electrical engineering one must calculate the right wire’s cross section based on the current this wire must handle. As a rule of thumb, we generally use a wire that can handle twice the current we want it to handle. 31.8.1

Types of Wires

There are basically two types of wires or cables in the market: stranded wire and solid core wire. Figure 31.9 shows a picture of both. Figure 31.9 Stranded and solid wire.

Solid core cables use one solid copper wire per conductor. It has a lower attenuation and is less costly than stranded cable; however it is designed for situations where they stay put without flexing once installed. The reason for that is that solid core wires are more rigid and flexing them constantly will make them break. Stranded cables consist of several strands of wires wrapped around each other in each conductor. They are much more flexible and consequently suited to applications that demand flexibility. Stranded cables have higher attenuation; therefore, they are better used over shorter distances.

31.9

Power Supply

Power supplies are another of the bench equipment. A bench power supply is a practical tool and allows circuits to be powered and consequently tested before they are finished.

31.12 Lead-Free Solder

This way, you do not need to have a specific power supply for each project you want to test, providing you with a reliable and stabilized source of power at different voltages and currents. Symmetrical power supplies are formed by three basic parts:

•• •

Display (A), where voltage and current levels are viewed. Rotary buttons (B) for adjusting voltage and current levels. Output jacks (C), where the selected voltage and current are delivered.

Notice that the output jacks provide symmetrical voltage. In the given example in Figure 31.10, the power supply is adjusted to provide 24 V and 2.3 A. At the output jacks, the power supply will provide −24 V, ground or 0 V, and +24 V, at the black (minus), green (ground), and red (positive) terminals, respectively.

Figure 31.10 Typical symmetrical power supply.

31.10

Soldering Station

A soldering station, mostly used in electronics and electrical engineering, consists of one or more soldering tools connected to the main unit, which includes temperature adjustment, holders and stands, soldering tip cleaners, etc. Soldering stations are widely used in electronics repair workshops or electronic laboratories or for household applications and hobbies.

31.11

Soldering Fume Extractors

Soldering produce fumes. Soldering fume extractors or absorbers are equipment that can extract or absorb these fumes generated by any soldering iron.

31.12

Lead-Free Solder

Solder generally contain lead that is a known neurotoxin and can pose other significant chronic health effects. Therefore, solder that contains lead is considered to be toxic.

517

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31 Instrumentation and Bench

The electronics industry has shifted to lead-free solder, but most hobbyists are still using lead in their solder. Traditional solder generally contains around 63% of tin and 27% of lead and melts at 183 C (361.4 F). Lead-free solder, on the other hand, contains around 96.5% of tin, 3% of silver, and 0.5% of copper and melts at 228 C (442.4 F), making it much harder to melt, requiring solder stations that can go that hot. Solder is sold in reels and can be either flux cored, with flux in the core of the wire, or solid, with no flux in the core. Cored solder wire is hollow solder wire with flux in the core. Solder flux is a kind of chemical used to remove any oxide, oil, grease, and other unwanted materials from the surfaces prior to soldering. Solder flux helps to deoxidize metals and helps better soldering and wetting. These impurities can affect the performance of circuits and cause future circuit failures. Therefore, using flux-cored solder saves both time and money.

31.13

A Few Images of Real Products

In the following images, nicely provided by Fluke and Weller, we show a few real equipment.

Figure 31.11 Fluke 87V (true RMS) multimeter.

31.13 A Few Images of Real Products

Figure 31.12 Fluke 190-204 scopemeter (portable oscilloscope).

From back to front in Figure 31.13, we see the temperature module, the iron support and cleaner, and the soldering iron.

Figure 31.13 Weller WE1010 soldering station.

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520

31 Instrumentation and Bench

The two elements with tubes in the background in Figure 31.14 are the fume extractor and the hot air soldering station.

Figure 31.14 Weller working bench.

Thanks Weller and Fluke for the images.

521

Appendix A International System of Units (SI)

Prefix

Symbol

Multiplier

Yotta

Y

1024 = 1000000000000000000000000

Zetta

Z

1021 = 1000000000000000000000

Exa

E

1018 = 1000000000000000000

Peta

P

1015 = 1000000000000000

Tera

T

1012 = 1000000000000

Giga

G

109 = 1000000000

Mega

M

106 = 1000000

Kilo

k

103 = 1000

Hecto

h

102 = 100

Deca

da

101 = 10

Deci

d

10−1 = 0.1

Centi

c

10−2 = 0.01

Milli

m

10−3 = 0.001

Micro

μ

10−6 = 0.000001

Nano

n

10−9 = 0.000000001 (Continued)

Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

522

Appendix A International System of Units (SI)

Prefix

Symbol

Multiplier

Pico

p

10−12 = 0.000000000001

Femto

f

10−15 = 0.000000000000001

Atto

a

10−18 = 0.000000000000000001

Zepto

z

10−21 = 0.000000000000000000001

Yocto

y

10−24 = 0.000000000000000000000001

523

Appendix B Color Code: Resistors

First stripe

Second stripe

Third stripe Gold

Black

Brown

Red

Orange

Yellow

Green

Blue

Brown

Black

1

10

100

1k

10k

100k

1M

10M

Brown

Brown

1.1

11

110

1.1k

11k

110k

1.1M

11M

Brown

Red

1.2

12

120

1.2k

12k

120k

1.2M

12M

Brown

Orange

1.3

13

130

1.3k

13k

130k

1.3M

13M

Brown

Green

1.5

15

150

1.5k

15k

150k

1.5M

15M

Brown

Blue

1.6

16

160

1.6k

16k

160k

1.6M

16M

Brown

Gray

1.8

18

180

1.8k

18k

180k

1.8M

18M

Red

Black

2

20

200

2k

20k

200k

2M

20M

Red

Red

2.2

22

220

2.2k

22k

220k

2.2M

22M

Red

Yellow

2.4

24

240

2.4k

24k

240k

2.4M

24M

Red

Violet

2.7

27

270

2.7k

27k

270k

2.7M

27M

Orange

Black

3

30

300

3k

30k

300k

3M

30M

Orange

Orange

3.3

33

330

3.3k

33k

330k

3.3M

33M

Orange

Blue

3.6

36

360

3.6k

36k

360k

3.6M

36M

(Continued) Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

524

Appendix B Color Code: Resistors

First stripe

Second stripe

Third stripe Gold

Black

Brown

Red

Orange

Yellow

Green

Blue

Orange

White

3.9

39

390

3.9k

39k

390k

3.9M

39M

Yellow

Orange

4.3

43

430

4.3k

43k

430k

4.3M

43M

Yellow

Violet

4.7

47

470

4.7k

47k

470k

4.7M

47M

Green

Brown

5.1

51

510

5.1k

51k

510k

5.1M

51M

Green

Blue

5.6

56

560

5.6k

56k

560k

5.6M

56M

Blue

Red

6.2

62

620

6.2k

62k

620k

6.2M

62M

Blue

Gray

6.8

68

680

6.8k

68k

680k

6.8M

68M

Violet

Green

7.5

75

750

7.5k

75k

750k

7.5M

75M

Gray

Red

8.2

82

820

8.2k

82k

820k

8.2M

82M

White

Brown

9.1

91

910

9.1k

91k

910k

9.1M

91M

525

Appendix C Root Mean Square (RMS) Value C.1

Introduction

In this appendix, we will demonstrate how the root mean square (RMS) value is derived in terms of infinitesimal calculus for the sinusoidal function.

C.2

RMS Value

A sinusoidal function is defined by the following generic equation. f t = A sin ωt + θ

C1

ω is the angular frequency, in radians per second. t is the time, in seconds. θ is the initial phase angle, in radians.1 A is the amplitude. where ω=

2π T

ω is the angular frequency, in radians per second. T is the period, in seconds. 1 In other words, the angle where the wave started. Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

C2

526

Appendix C Root Mean Square (RMS) Value

In statistics, the RMS value of any function is given by the following equation: 1 T

VRMS =

T

f t

2

dt

0

We can simplify this equation by squaring both sides: V2RMS =

1 T

T

f t

2

dt

0

If we substitute f(t) by the sinusoidal function (C.1), we get V2RMS =

1 T

T

A sin ωt + θ

2

dt

0

Solving… T

V2RMS =

1 T

V2RMS =

1 2 A T

A2 sin2 ωt + θ dt

0 T

sin2 ωt + θ dt

0

For the purposes of this calculation, we can disregard θ by making it 0: V2RMS =

1 2 A T

V2RMS =

1 2 A T

T 0 T

sin2 ωt + 0 dt sin2 ωt dt

0

MATH CONCEPT sin2 θ = θ2 − 14 sin 2θ

If we apply this concept, we obtain V2RMS

1 t 1 = A2 − sin 2ωt T 2 4

T 0

Solving for the interval, V2RMS =

1 2 A T

T 1 − sin 2ωT 2 4

−

ω0 1 − sin 2ω 0 4 2

Appendix C Root Mean Square (RMS) Value

V2RMS =

1 2 A T

T 1 − sin 2ωT 2 4

1 − 0 − sin 0 4

T 1 − 0 2 4

1 0 4

MATH CONCEPT sin 0 = sin xπ = 0

V2RMS =

1 2 A T

V2RMS =

1 2 T A T 2

Simplifying T, V2RMS = VRMS =

A2 2 A2 2

RMS VALUE VRMS =

A 2

VRMS is the RMS value. A is the amplitude value.

− 0−

527

529

Appendix D Complex Numbers D.1

Real Numbers

Real numbers can be seen as numbers that exist in a horizontal line. Zero is at the center. Positive numbers extend to the right and negative numbers to the left, like shown in Figure D.1.

Figure D.1 Real numbers.

D.2

Complex Numbers

A complex number is represented by two axes, a real (horizontal) and an imaginary (vertical) one, like shown in Figure D.2. This characteristic of complex numbers provides these numbers with the ability to express three quantities: two numerical values and an angle. An imaginary number has the following form, called a rectangular form. a + bi

where a is the real part and b the imaginary part. The imaginary part is represented with a letter, generally i or j.

Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

530

Appendix D Complex Numbers

Figure D.2 Imaginary number axes.

Figure D.3 Imaginary number example.

3 + 2i is an imaginary number that can be represented on the complex number plane as seen in Figure D.3. D.2.1

Operations with Complex Numbers

D.2.1.1 Addition

To add two complex numbers, the real and the imaginary parts must be added alone.

Appendix D Complex Numbers

Therefore, +

3 + 2i 5 − 10i 8 − 8i

D.2.1.2 Subtraction

To subtract two complex numbers, the real and the imaginary parts must be subtracted alone. Therefore, −

13 + 12i 15 −100i − 2 − 88i

D.2.1.3 Multiplication

The product of two complex numbers follows the distributive law. The multiplication of a + bi and c + di will be ac + a di + bi c + bi di Therefore, 7 + 10i × 4 −9i will be 7 × 4 + 7 × −9i + 10i × 4 + 10i × − 9i 28−63i + 40i−90i2

MATH CONCEPT i2 = − 1

Therefore, 28−23i −90 −1 28−23i + 90

FINAL RESULT 118 −23i

531

532

Appendix D Complex Numbers

D.2.1.4 Division

The division of two complex numbers is made by multiplying both numbers by the conjugate of the second number. The conjugate of a complex number is the same number with the imaginary part taken with the inverse sign. The conjugate of a + bi is a − bi. Thus, 4 + 2i 2 −i 4 + 2i 2 + i 2−i 2 + i 4 × 2 + 4 × i + 2i × 2 + 2i × i 2 × 2 + 2 × i + −i × 2 + −i × i 8 + 4i + 4i + 2i2 4 + 2i−2i− i2

MATH CONCEPT i2 = − 1 8 + 4i + 4i + 2 −1 4 + 2i−2i− −1 8 + 4i + 4i−2 4 + 2i−2i + 1

FINAL RESULT 6 + 8i 5

D.3

Polar Numbers

By observing Figure D.3, it is clear that number 3 + 2i can be seen as a vector with a certain magnitude R and an angle θ with the real axis, as shown in Figure D.4.

Appendix D Complex Numbers

Figure D.4 Imaginary number represented as a vector.

Figure D.5 Imaginary number – real and imaginary parts.

By trigonometry, the vector can be decomposed into real RRE and imaginary RIM parts as illustrated in Figure D.5. REAL COMPONENT RRE = R cos θ RRE is the real component. θ is the angle of R with the real axis.

IMAGINARY COMPONENT RIM = R sin θ RIM is the imaginary component. θ is the angle of R with the real axis.

533

534

Appendix D Complex Numbers

Therefore, the magnitude can be found by the following formula. MAGNITUDE R=

R2RE + R2IM

R is the magnitude. RRE is the real component. RIM is the imaginary component.

where

θ = A tan

RIM RRE

This formula will always be true for angles, one positive and one negative, meaning two angles separated by 180 , or π. The best way to always find the correct value for θ is to use one of the two following formulas. USE THIS FORMULA IF THE REAL PART IS NEGATIVE θ = A tan

RIM +π RRE

θ is the angle of R with the real axis. RRE is the real component. RIM is the imaginary component.

USE THIS FORMULA IF THE REAL PART IS POSITIVE θ = A tan

RIM RRE

θ is the angle of R with the real axis. RRE is the real component. RIM is the imaginary component.

Modern calculators have the function ATAN2 that calculates the angle correctly, because they take into consideration the real part sign.

Appendix D Complex Numbers

Therefore, complex numbers can also be expressed in phasor form. FORMULA Z=

R2RE + R2IM

∠θ

Z is the number in polar form. θ is the angle of R with the real axis. RRE is the real component. RIM is the imaginary component.

D.4

Trigonometric Numbers

Complex numbers can also be expressed as trigonometric numbers in the following form. FÓRMULA Z = R cos θ + i sin θ Z is the trigonometric number. R is the magnitude. θ is the angle of R with the real axis.

535

537

Appendix E Table of Integrals

Integral

Result

xndx

1 n+1 x n+1

1 x dx

ln(x) x > 0 − ln(x) x < 0

c dx ax + b

c ln ax + b a

1 x+a

2 dx

1 dx a −x (x + a)ndx

−

1 x+a

2 a− x

x+a

n

if x

a x + 1+x 1+n −1

ln(x)dx

x ln(x) − x

ln(ax)dx

x ln(ax) − x

ln ax dx x

1 ln ax 2

eaxdx

1 ax e a

xexdx

(x − 1)ex

2

(Continued)

Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

538

Appendix E Table of Integrals

Integral

Result

xeaxdx

x 1 ax − e a a2

xe −ax dx

−

ex sin (x)dx

1 x e sin x − cos x 2

sin (x)dx

− cos (x)

cos (x)dx

sin(x)

tan (x)dx

ln sec (x)

tan (x)dx

− ln [cos (x)]

x sin(x)dx

−x cos(x) + sin (x)

x cos(x)dx

cos (x) + x sin(x)

sin (x) cos (x)dx

−

sin2(x)dx

x 1 − sin 2x 2 4

cos2(x)dx

x 1 + sin 2x 2 4

sinh (x)dx

cosh (x)

cosh (x)dx

sinh (x)

2

1 − ax2 e 2a

1 cos x 2

2

539

Appendix F AWG Versus Metric System: Wire Cross Sections

Metric system (mm2)

Diameter (inches)

Diameter (mm)

Cross section (mm2)

0000 (4/0)

0.460

11.7

000 (3/0)

0.410

10.4

00 (2/0)

0.365

9.27

67.4

0 (1/0)

0.325

8.25

53.5

1

0.289

7.35

42.4

2

0.258

6.54

33.6

3

0.229

5.83

26.7

4

0.204

5.19

21.1

5

0.182

4.62

16.8

6

0.162

4.11

13.3

7

0.144

3.67

10.6

8

0.129

3.26

8.36

9

0.114

2.91

6.63

10

0.102

2.59

5.26

11

0.0907

2.30

4.17

107 85.0

(Continued)

Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

540

Appendix F AWG Versus Metric System: Wire Cross Sections

Metric system (mm2)

Diameter (inches)

Diameter (mm)

Cross section (mm2)

12

0.0808

2.05

3.31

13

0.0720

1.83

2.63

14

0.0641

1.63

2.08

15

0.0571

1.45

1.65

16

0.0508

1.29

1.31

17

0.0453

1.15

1.04

18

0.0403

1.02

0.82

19

0.0359

0.91

0.65

20

0.0320

0.81

0.52

21

0.0285

0.72

0.41

22

0.0254

0.65

0.33

23

0.0226

0.57

0.26

24

0.0201

0.51

0.20

25

0.0179

0.45

0.16

26

0.0159

0.40

0.13

X/0 is used in the American Wire Gauge (AWG) system for wires with diameters greater than 1. The number X represents the number of zeros and pronounced like “aught.” 3/0, for example, pronounced “three aught” means “three zeros,” or AWG 000.

541

Appendix G Resistors: Commercial Values

0.1%, 0.25%, and 0.5% resistors

10

10.1

10.2

10.4

10.5

10.6

10.7

10.9

11

11.1

11.3

11.4

11.5

11.7

11.8

12

12.1

12.3

12.4

12.6

12.7

12.9

13

13.2

13.3

13.5

13.7

13.8

14

14.2

14.3

14.5

14.7

14.9

15

15.2

15.4

15.6

15.8

16

16.2

16.4

16.5

16.7

16.9

17.2

17.4

17.6

17.8

18

18.2

18.4

18.7

18.9

19.1

19.3

19.6

19.8

20

20.3

20.5

20.8

21

21.3

21.5

21.8

22.1

22.3

22.6

22.9

23.2

24.4

23.7

24

24.3

24.6

24.9

25.2

25.5

25.8

26.1

26.4

26.7

27.1

27.4

27.7

28

28.4

28.7

29.1

29.4

29.8

30.1

30.5

30.9

31.2

31.6

32

32.4

32.8

33.2

33.6

34

34.4

34.8

35.2

35.7

36.1

36.5

37

37.4

37.9

38.3

38.8

39.2

39.7

40.2

40.7

41.2

41.7

42.2

42.7

43.2

43.7

44.2

44.8

45.3

45.9

46.4

47

47.5

48.1

48.7

49.3

49.9

50.5

51.1

51.7

52.3

53

53.6

54.2

54.9

55.6

56.2

56.9

57.6

58.3

59

59.7

60.4

61.2

61.9

62.9

63.4

64.2

64.9

65.7

65.5

67.3

68.1

69

69.8

70.6

71.5

72.3

73.2

74.1

75

75.9

76.8

77.7

78.7

79.6

80.6

81.6

82.5

83.5

84.5

85.6

86.6

87.6

88.7

89.8

90.9

92

93.1

94.2

95.3

96.5

97.6

98.8

Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

542

Appendix G Resistors: Commercial Values

1% resistors

10

10.2

10.5

10.7

11

11.3

11.5

11.8

12.1

12.4

12.7

13

13.3

13.7

14

14.3

14.7

15

15.4

15.8

16.2

16.5

16.9

17.4

17.8

18.2

18.7

19.1

19.6

20

20.5

21

21.5

22.1

22.6

23.2

23.7

24.3

24.9

25.5

26.1

26.7

27.4

28

28.7

29.4

30.1

30.9

31.6

32.4

33.2

34

34.8

35.7

36.5

37.4

38.3

39.2

40.2

41.2

42.2

43.2

44.2

45.3

46.4

47.5

48.7

49.9

51.1

52.3

53.6

54.9

56.2

57.6

59

60.4

61.9

63.4

64.9

66.5

68.1

69.8

71.5

73.2

75

76.8

78.7

80.6

82.5

84.5

86.6

88.7

90.9

93.1

95.3

97.6

5% resistors

10

11

12

13

15

16

18

20

22

24

27

30

33

36

39

43

47

51

56

62

68

75

82

91

10% resistors

10

12

15

18

22

27

33

39

47

56

68

82

100

120

150

180

220

270

330

390

470

560

680

820

1000

1200

1500

1800

2200

2700

3300

3900

4700

5600

6800

8200

10000 12000 15000 18000 22000

27000 33000 39000 47000 56000 68000 82000

100000 120000 150000 180000 220000 270000 330000 390000 470000 560000 680000 820000 20% resistors

10

15

22

33

47

68

100

150

220

330

470

680

1000

1500

2200

3300

4700

6800

10000

15000

22000

33000

47000

68000

100000

150000

220000

330000

470000

680000

543

Appendix H Capacitors: Commercial Values

Electrolytic capacitors

0.1 μF

12 μF

60 μF

220 μF

460 μF

1000 μF

3500 μF

7400 μF

25000 μF

62000 μF

0.15 μF

15 μF

68 μF

230 μF

470 μF

1100 μF

3600 μF

7600 μF

26000 μF

66000 μF

0.22 μF

16 μF

72 μF

233 μF

480 μF

1200 μF

3700 μF

7800 μF

27000 μF

68000 μF

0.33 μF

18 μF

75 μF

240 μF

500 μF

1300 μF

3900 μF

8200 μF

28000 μF

76000 μF

0.47 μF

20 μF

82 μF

243 μF

510 μF

1400 μF

4000 μF

8300 μF

30000 μF

0.1 F

0.68 μF

21 μF

88 μF

250 μF

520 μF

1500 μF

4100 μF

8400 μF

31000 μF

0.11 F

1 μF

22 μF

100 μF

270 μF

540 μF

1600 μF

4200 μF

8700 μF

32000 μF

0.12 F

1.5 μF

24 μF

108 μF

300 μF

550 μF

1700 μF

4300 μF

9000 μF

33000 μF

0.15 F

2 μF

25 μF

120 μF

320 μF

560 μF

1800 μF

4600 μF

9600 μF

34000 μF

0.22 F

2.2 μF

27 μF

124 μF

324 μF

590 μF

2000 μF

4700 μF

10000 μF

36000 μF

0.33 F

(Continued)

Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

544

Appendix H Capacitors: Commercial Values

Electrolytic capacitors

3 μF

30 μF

130 μF

330 μF

620 μF

2100 μF

4800 μF

11000 μF

37000 μF

0.47 F

3.3 μF

33 μF

140 μF

340 μF

645 μF

2200 μF

5000 μF

12000 μF

38000 μF

0.666 F

4 μF

35 μF

145 μF

350 μF

650 μF

2500 μF

5100 μF

13000 μF

39000 μF

4.7 μF

36 μF

150 μF

370 μF

680 μF

2600 μF

5400 μF

15000 μF

40000 μF

5 μF

39 μF

161 μF

378 μF

700 μF

2700 μF

5500 μF

16000 μF

41000 μF

5.6 μF

40 μF

170 μF

380 μF

708 μF

2800 μF

5600 μF

17000 μF

47000 μF

6.8 μF

43 μF

180 μF

390 μF

730 μF

2900 μF

5800 μF

18000 μF

48000 μF

7 μF

47 μF

189 μF

400 μF

800 μF

3000 μF

6000 μF

20000 μF

50000 μF

8 μF

50 μF

200 μF

420 μF

820 μF

3100 μF

6500 μF

22000 μF

55000 μF

8.2 μF

53 μF

210 μF

430 μF

850 μF

3300 μF

6800 μF

23000 μF

56000 μF

10 μF

56 μF

216 μF

450 μF

860 μF

3400 μF

7200 μF

24000 μF

60000 μF

Ceramic and other types of capacitors μF

nF

pF

Code

μF

nF

pF

100K

0.00001

0.01

10

339K

0.0000033

0.0033

3.3

101K

0.0001

0.1

100

390K

0.000039

39

39

102K

1

1

1000

391K

0.00039

0.39

390

102K

1

1

1000

392K

0.0039

3.9

3900

103K

0.01

10

10000

393K

39

39

39000

104K

0.1

100

100000

394K

0.39

390

390000

Code

Appendix H Capacitors: Commercial Values

Ceramic and other types of capacitors μF

nF

pF

Code

μF

nF

pF

105K

1

1000

1000000

399K

0.0000039

0.0039

3.9

109K

0.000001

1

1

400K

0.00004

0.04

40

120K

0.000012

12

12

401K

0.0004

0.4

400

121K

0.00012

0.12

120

402K

4

4

4000

122K

0.0012

1.2

1200

403K

0.04

40

40000

123K

12

12

12000

404K

0.4

400

400000

124K

0.12

120

120000

409K

0.000004

4

4

129K

0.0000012

0.0012

1.2

470K

0.000047

47

47

150K

0.000015

15

15

471K

0.00047

0.47

470

151K

0.00015

0.15

150

472K

0.0047

4.7

4700

152K

0.0015

1.5

1500

473K

47

47

47000

153K

15

15

15000

474K

0.47

470

470000

154K

0.15

150

150000

479K

0.0000047

0.0047

4.7

159K

0.0000015

0.0015

1.5

500K

0.00005

0.05

50

180K

0.000018

18

18

501K

0.0005

0.5

500

181K

0.00018

0.18

180

502K

5

5

5000

182K

0.0018

1.8

1800

503K

0.05

50

50000

183K

18

18

18000

504K

0.5

500

500000

184K

0.18

180

180000

509K

0.000005

5

5

189K

0.0000018

0.0018

1.8

560K

0.000056

56

56

200K

0.00002

0.02

20

561K

0.00056

0.56

560

201K

0.0002

0.2

200

562K

0.0056

5.6

5600

Code

(Continued)

545

546

Appendix H Capacitors: Commercial Values

Ceramic and other types of capacitors μF

nF

pF

Code

μF

nF

pF

202K

2

2

2000

563K

56

56

56000

203K

0.02

20

20000

564K

0.56

560

560000

204K

0.2

200

200000

569K

0.0000056

0.0056

5.6

209K

0.000002

2

2

600K

0.00006

0.06

60

220K

0.000022

22

22

601K

0.0006

0.6

600

221K

0.00022

0.22

220

602K

6

6

6000

222K

0.0022

2.2

2200

603K

0.06

60

60000

223K

22

22

22000

604K

0.6

600

600000

224K

0.22

220

220000

609K

0.000006

6

6

229K

0.0000022

0.0022

2.2

680K

0.000068

68

68

250K

0.000025

25

25

681K

0.00068

0.68

680

251K

0.00025

0.25

250

682K

0.0068

6.8

6800

252K

0.0025

2.5

2500

683K

68

68

68000

253K

25

25

25000

684K

0.68

680

680000

254K

0.25

250

250000

689K

0.0000068

0.0068

6.8

259K

0.0000025

0.0025

2.5

700K

0.00007

0.07

70

270K

0.000027

27

27

701K

0.0007

0.7

700

271K

0.00027

0.27

270

702K

7

7

7000

272K

0.0027

2.7

2700

703K

0.07

70

70000

273K

27

27

27000

704K

0.7

700

700000

274K

0.27

270

270000

709K

0.000007

7

7

279K

0.0000027

0.0027

2.7

800K

0.00008

0.08

80

300K

0.00003

0.03

30

801K

0.0008

0.8

800

Code

Appendix H Capacitors: Commercial Values

Ceramic and other types of capacitors μF

nF

pF

Code

μF

nF

pF

301K

0.0003

0.3

300

802K

8

8

8000

302K

3

3

3000

803K

0.08

80

80000

303K

0.03

30

30000

804K

0.8

800

800000

304K

0.3

300

300000

809K

0.000008

8

8

309K

0.000003

3

3

820K

0.000082

82

82

330K

0.000033

33

33

821K

0.00082

0.82

820

331K

0.00033

0.33

330

822K

0.0082

8.2

8200

332K

0.0033

3.3

3300

823K

82

82

82000

333K

33

33

33000

824K

0.82

820

820000

334K

0.33

330

330000

829K

0.0000082

0.0082

8.2

Code

547

549

Appendix I Inductors: Commercial Values

Inductors (pH)

100

500

150

560

180

600

200

670

220

680

270

700

300

780

330

800

390

820

400

900

470

Inductors (nH)

1

4.7

8.7

20

50

93

157

247

380

590

1.1

4.8

8.8

20.8

51

94

160

250

386

592

1.15

4.9

8.9

21

52

95

163

252

387

600

(Continued) Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

550

Appendix I Inductors: Commercial Values

Inductors (nH)

1.2

5

9

21.5

53

96

164

256

390

604

1.3

5.1

9.1

21.7

54

97

165

257

397

610

1.4

5.2

9.2

22

55

98

166

260

400

620

1.5

5.3

9.3

23

55.5

98.5

169

262

405

624

1.6

5.4

9.4

23.5

56

98.7

170

264

410

630

1.65

5.45

9.5

24

58

99

175

269

416

640

1.7

5.5

9.6

25

59

100

178

270

420

650

1.8

5.6

9.7

26

59.5

101

179.6

272

422

660

1.9

5.7

9.8

26.5

60

102

180

276

430

670

2

5.8

9.85

27

61

105

182

280

434

674

2.1

5.9

9.9

27.3

61.5

106

185

282

435

680

2.2

6

10

28

62

108

187

285

440

690

2.3

6.1

10.2

29

64

110

188

288

442

700

2.4

6.2

10.4

30

65

111

190

290

450

708

2.5

6.3

11

31

66

114

191

292

460

709

2.55

6.4

11.2

32

67

115

192

298

463

720

2.6

6.5

11.9

32.75

67.8

117

194

300

465

730

2.7

6.6

12

33

68

119.7

195

307

468

740

2.8

6.68

12.1

34

69

120

196

310

470

750

2.9

6.7

12.3

35

70

122

197

311

477

760

3

6.8

12.5

35.5

71

123

200

315

480

764

3.1

6.9

12.55

36

72

125

202

317

490

770

3.2

7

13

37

73

126

205

320

491

780

3.3

7.1

13.7

37.5

74

130

206

322

500

783

3.32

7.15

13.8

38.5

75

132

207

325

505

790

Appendix I Inductors: Commercial Values

Inductors (nH)

3.4

7.2

14

39

76

134

210

330

509

800

3.5

7.3

14.7

40

77

135

215

331

510

814

3.6

7.4

14.9

41

78

139

216

333

520

820

3.7

7.5

15

42

79

140

217

337

523

845

3.8

7.6

15.7

43

80

141

220

340

530

846

3.85

7.7

16

43.5

82

144

222

342

538

850

3.9

7.8

16.6

44

82.7

145

223

344

540

870

4

7.9

17

44.5

83

146

225

350

545

880

4.1

8

17.5

45

84

149

226

360

548

896

4.2

8.1

17.8

46

85

149.4

230

363

550

900

4.3

8.2

17.95

46.5

86

150

240

365

556

908

4.4

8.3

18

46.6

88

152

241

369

560

909.5

4.5

8.4

18.5

47

90

154

242

370

570

910

4.55

8.5

19

48

91

155

245

375

571

950

4.6

8.6

19.4

49

92

156

246

378

580

978

Inductors (μH)

1

3.3

7.8

12.2

22

38.07

69.7

125

252

521

1.01

3.32

7.8

12.3

22.3

38.3

69.77

128

253

530

1.02

3.35

7.9

12.5

22.39

39

70

130

256

546

1.06

3.4

7.94

12.5

22.5

39.5

70.05

131

258

550

1.08

3.5

8

12.6

22.6

39.9

70.5

131.8

260

560

1.09

3.6

8

12.7

22.7

40

71.1

132

262

561

1.1

3.7

8.06

12.8

22.8

41

73

133.2

264.7

562

(Continued)

551

552

Appendix I Inductors: Commercial Values

Inductors (μH)

1.15

3.76

8.1

12.9

23

41.1

73.7

134.2

270

570

1.18

3.8

8.17

13

23.3

41.5

75

137.5

272

575

1.2

3.9

8.2

13.1

23.7

42

76.8

140

275

579.4

1.23

4

8.23

13.2

24

43

77

143

285

580

1.25

4.1

8.3

13.3

24.

43.3

77.5

143.5

290

590

1.26

4.2

8.32

13.5

24.2

44

78

145

296.7

595.7

1.27

4.23

8.4

13.9

24.31

45

78.4

146

300

600

1.3

4.3

8.45

14

24.4

46

79

148

303.3

602.5

1.32

4.4

8.5

14.2

25

46.3

80

148.5

305

620

1.33

4.5

8.6

14.3

25.7

46.7

80.75

150

307

620

1.4

4.6

8.7

14.6

26

46.8

82

151

316

630

1.45

4.7

8.8

14.67

26.2

47

82.2

156

320

640

1.47

4.74

8.8

14.7

27

47.1

83

159.65

324

645

1.5

4.76

8.83

14.75

27.1

47.4

84

160

325

648

1.58

4.8

8.84

14.8

27.16

47.7

85

160.6

325

650

1.6

4.9

8.86

14.9

27.4

49.7

86

165

325.1

656.2

1.65

4.95

8.9

15

27.9

50

87

166

330

670

1.68

5

9

15.14

28

51

88

167

332

680

1.7

5.1

9.1

15.4

28.2

51.2

88.2

167.2

335

682.8

1.75

5.2

9.2

15.5

28.4

51.9

89.5

168

350

696

1.7929

5.3

9.3

15.8

29

52

89.6

170

357

700

1.8

5.4

9.4

15.9

29.1

52.5

89.7

173

360

749

1.83

5.5

9.5

16

29.5

52.6

90

175

370

750

1.84

5.6

9.6

16.1

29.7

53

91

176

372

760

Appendix I Inductors: Commercial Values

Inductors (μH)

1.87

5.7

9.62

16.2

29.8

54

92

176.1

374

770

1.875

5.78

9.65

16.3

30

54.2

92.5

176.4

375

780

1.9

5.8

9.67

16.5

30.6

54.4

93

178

377

781

1.98

5.9

9.7

16.6

31

54.81

95

180

380

786

2

6

9.8

16.7

31.8

55

96

181

383

790

2.02

6.1

9.9

16.76

31.84

55.2

97

187.4

390

800

2.03

6.2

10

16.8

32

55.23

97.7

190

399

807

2.1

6.23

10

17

32.5

55.4

98

192

400

810

2.15

6.3

10.1

17.6

32.6

55.6

98.8

197

402

820

2.16

6.4

10.2

18

32.8

56

99.3

200

417.4

830

2.17

6.4

10.27

18.2

32.9

57

99.5

201.6

420

832

2.2

6.5

10.4

18.3

33

57.8

100

203.7

426.3

836

2.3

6.6

10.5

18.5

33.15

58

101

205

429.6

849

2.32

6.63

10.6

18.64

33.2

58.1

101.7

207.2

430

850

2.4

6.7

10.68

18.85

33.3

60

102

209.6

439

860

2.42

6.8

10.7

19

33.33

61

105

210

440

880

2.45

6.9

10.79

19.4

33.7

61.9

107.5

211

450

900

2.5

7

10.9

19.5

35

62

110

216.7

450

910

2.54

7.03

11

19.7

35.2

63.2

111.6

217

458.7

915

2.6

7.1

11.1

20

35.3

64

112

218.8

460

940

2.7

7.2

11.2

20.5

35.48

64.2

113.2

220

470

950

2.76

7.22

11.3

20.8

35.5

65

114

223

471

955

2.8

7.28

11.3

20.9

35.6

65.04

115

224

472

960 (Continued)

553

554

Appendix I Inductors: Commercial Values

Inductors (μH)

2.84

7.3

11.4

21

36

66.5

116

225

476

2.9

7.4

11.5

21.47

36.6

66.89

118

230

480

3

7.5

11.6

21.5

37

67

120

231

497

3.05

7.5

11.8

21.6

37.5

67.6

121

240

500

3.1

7.6

11.9

21.65

37.6

68

122.4

245.8

501

3.2

7.63

12

21.8

37.9

68.22

122.5

248

510

3.22

7.7

12

21.9

38

68.37

123

250

518

980

Inductors (mH)

1

1.8

3.1

5.3

8.3

13.2

21.5

37.8

61

142

1

1.82

3.2

5.4

8.4

13.4

22

38

62

150

1.0017

1.9

3.3

5.5

8.8

13.6

23

39

65.6

158

1.008

1.95

3.4

5.6

9

14

23.7

40

66

180

1.03

1.97

3.5

5.7

9.1

14.5

24

40.5

67.5

190

1.05

2

3.6

5.85

9.2

14.6

24.5

40.8

68

200

1.07

2.1

3.7

5.9

9.3

14.8

25

41

70

208

1.08

2.2

3.8

6

9.35

15

25.5

41.5

70.3

220

1.1

2.2

3.9

6.1

9.4

15.1

26

42

75

250

1.134

2.204

4

6.2

9.5

15.4

26.2

42.5

76

265

1.17

2.25

4.1

6.4

9.6

15.5

26.5

43.6

77

270

1.186

2.3

4.15

6.5

9.8

16

26.7

44

78

300

1.2

2.35

4.2

6.6

9.9

16.2

27

45

82

320

1.203

2.36

4.25

6.7

10

16.4

27.7

46.7

82.5

330

1.25

2.37

4.3

6.8

10.2

16.9

28

47

90

370

1.3

2.38

4.4

6.9

10.5

17

28.4

48

91

500

Appendix I Inductors: Commercial Values

Inductors (mH)

1.31

2.4

4.5

7

10.8

17.1

28.6

49

92

560

1.32

2.48

4.6

7.1

10.9

17.7

29

50

92.5

600

1.34

2.5

4.66

7.2

11

17.8

29.6

50.7

93

680

1.4

2.55

4.7

7.3

11.2

18

30

51

100

900

1.47

2.6

4.75

7.4

11.4

18.4

32

52

105

1.48

2.61

4.8

7.5

11.5

18.5

32.4

53.5

110

1.499

2.7

4.81

7.6

12

18.52

33

54

115.7

1.5

2.8

4.9

7.7

12.2

19

34

55

120

1.58

2.88

4.91

7.8

12.3

19.4

35

56

125

1.6

2.89

5

8

12.5

19.6

36

57

135

1.62

2.9

5.1

8.1

12.8

20

37

59.6

138

1.7

3

5.2

8.2

13

21.2

37.7

60

140

Inductors (H)

1

3

8

20

1.5

3.5

9

30

2

4

10

40

2.2

5

12

60

2.5

6

14

150

2.6

7

15

555

557

Appendix J Simulation Tools The lists that follow show the best simulation tools you will find around, in our opinion, that can help you create, test, and simulate electronic devices. The list is in alphabetic order. The bold ones are the ones we prefer: Online simulators www.circuitlab.com www.everycircuit.com www.partsim.com/simulator www.systemvision.com Software (Windows and Mac) www.analog.com/en/design-center/design-tools-and-calculators/ltspice-simulator.html www.autodesk.com/products/eagle/free-download www.circuitmaker.com

Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

559

Appendix K Glossary

Glossary

Ω

Omega is the symbol used to express resistance, in Ohms.

A

Ampere is the unit used to express current.

AC

Alternating current is a kind of current that has its value changing cyclically over time according to a sinusoidal function.

Admittance

In electrical engineering, admittance is a measure of how easily a circuit or device will allow a current to flow. It is defined as the reciprocal of impedance.

Anode

An anode is the positively charged electrode of a device.

C

Coulomb is the unit used to express electric charge.

Capacitance

Capacitance is the ability of a body to store potential energy in the form of electrical charge, electrical field.

Capacitor

Is a passive device that can store potential energy in the form or electric field. See capacitance.

Cathode

A cathode is the negatively charged electrode of a device.

DC

Direct current is a kind of current that keeps its value constant over time.

DC bias

See DC offset.

DC offset

Is a mean amplitude displacement in AC voltage by the existence or a spurious DC component added to it. (Continued)

Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

560

Appendix K Glossary

Glossary

Dielectric

A dielectric is an electrical insulator that can be polarized by an applied electric field.

Diode

Is a semiconductor device that can conduct current in one direction only.

Electrolytic capacitor

Is a polarized capacitor that can, among other things, have high capacitance values without increasing too much in size. See capacitor.

F

Farad is the unit used to express capacitance. See capacitor.

g

Gram is the unit used to express mass.

H

Henry is the unit used to express inductance. See inductor.

Hz

Hertz is the unit used to express frequency.

Impedance

The effective resistance of an electric circuit or component to alternating current, arising from the combined effects of ohmic resistance and reactance.

Inductance

Inductance is the ability of a body to store potential energy in the form of magnetic field. See inductor.

Inductor

Is a passive device that can store potential energy in the form or magnetic field. See inductance.

J

Joule is the unit used to express energy.

m

Meter is the unit used to express distance.

N

Newton is the unit used to express force.

Permittivity

Permittivity is the measure of a material’s ability to store an electric field in the polarization of the medium.

Phase

Is a difference in angle between two alternating signals.

Potentiometer

Is a three terminal variable resistor. See resistor.

Resistance

Is the property of a body that opposes the flow of current.

Resistor

Is a passive device that resists current flow.

S

Siemens is the unit used to express admittance.

Transformer

Is a device that can be used to reduce or to increase the voltage of an alternating current signal.

Appendix K Glossary

Glossary

Transistor

Is an active semiconductor device that can be used to perform many functions like amplification, rectification, switching, etc.

Trimmer

Is a miniature adjustable component that can be a variable potentiometer, a variable capacitor, or a variable inductor. See resistor, capacitor, and inductor.

Trimpot

Is a miniature adjustable potentiometer. See potentiometer.

V

Volt is the unit used to express electric potential.

VA

Volt-ampere is the unit used to express apparent power.

VAR

Volt-ampere reactive is the unit used to express reactive power.

W

Watt is the unit used to express real or active power.

561

563

Index a alternating current (AC) 34 analysis 463 characteristics 34 cycles 34 motors 132 peak-to-peak voltage 36 period and frequency 35 voltage sources, independent 147 alternating power 184 American wire gauges (AWG) 539 ammeter 511 amplifiers 451 classes 454 apparent power 326 atoms and quarks 19 audio signal 463 average value 185

b barrier voltage 140 batteries in parallel 29 in series 29 short-circuiting 30 warnings 30 battery life 28 bench equipment ammeter 511 breadboards 513

images of real products 519 lead-free solder 519 ohmmeter 512 oscilloscope 513 power supply 517 root mean square (RMS) multimeters 509 soldering fume extractors 517 soldering station 517 voltmeter 510 β 452 biasing 144, 455 curve 138 forward 138 reverse 138 branch currents 204 breadboards 513

c calculus 9 capacitance 78 capacitive impedance 167 capacitive reactance 88 capacitors 73 AC analysis 87 capacitance 78 capacitive reactance 88 color code 95 commercial values 543 construction methods 76

Introductory Electrical Engineering with Math Explained in Accessible Language, First Edition. Magno Urbano. © 2020 John Wiley & Sons, Inc. Published 2020 by John Wiley & Sons, Inc.

564

Index

capacitors (cont’d) current on a charging 81 dielectric 75 electric characteristics 77 electric field 78 electrolytic 91 example 1 84 example 2 86 exercises 98 history 73 how it works 73 markings 96 in parallel 94 phase 88 pool effect 87 in series 93 stored energy 79 variable 93 voltage and current 81 voltage on charging 82 circuit analysis 215, 216, 235, 257, 380 class A amplifier 454 coils and magnets 110 collector current 459, 460 collector feedback biasing 466 color code, resistors 523 common emitter 455 complex numbers 529 addition 530 division 532 multiplication 531 subtraction 531 conductors and electricity 22 conventions xxv cross sections 539 current 25 across an inductor 116 on charging capacitor 81 gain 452 sources dependent 149 independent 149

d Danger Will Robison, Danger! 30 Delta circuit 63 Delta–Wye conversion 63 example 1 65 example 2 67 exercise 1 69 exercise 2 70 dependent current sources 149 dependent voltage sources 148 derivatives 11 differentiator (operational amplifiers) 507 diodes 143 direct current (DC) 31 characteristics 31 motors 131 offset 37 voltage sources, independent 147

e electric field 78 electric generators 129 how they work 129 electricity, basic rules 51 electric motor 131 AC 132 DC 131 electric potential 24 electric power and work 181 electric resistance 25, 41 electric schematics xxv electrolytic capacitor 91 electromagnetism 103 electromotive force 116 electrons 20 shells 23 examples capacitors 84, 86 Delta configuration 65 fake 67 impedance and phase 169, 171, 173

Index

Kirchhoff’s laws 199, 205 Millman’s theorem 291, 293 nodal analysis 215, 219, 221 Norton’s theorem 258 Ohm’s laws 53, 55, 56, 57 RC circuits 315, 317, 320, 322 RLC circuits–current analysis 432, 437 RLC circuits–voltage analysis 409, 412 RL circuits 354, 355, 358 source transformations 153 superposition theorem 271 Thévenin theorem 236 exercises Delta–Wye configuration 69 impedance and phase 177 inductors 120 Kirchhoff’s laws 210 Millman’s theorem 293, 294 nodal analysis 226 Norton’s theorem 253 Ohm’s laws 58 RC circuits 328 resistors 48 RLC circuits–current analysis 442 RLC circuits–voltage analysis 418 RL circuits 362 source transformations 160 superposition theorem 281 Thévenin theorem 250 transistor amplifiers 477 voltage and current 38

f fake Delta configuration 67 Faraday first experiment 105 law, transformers 124 second experiment 106 first oven 185 fixed base biasing 455 forward biasing 138

g ground

33

h Hans Christian Ørsted 103 hybrid parameter, forward transfer, and common emitter (hFE) 452

i impedance 166, 177, 323 capacitive 167 example 1 169 example 2 171 importance in real life example 173 inductive 169 parallel 166 and phase 165, 177 exercises 177 series 166 what about a bigger load? 175 independent AC voltage sources 147 independent current sources 149 independent DC voltage sources 147 inductance 111 inductive impedance 169 inductors 109 AC analysis 116 alternating voltage and current 117 commercial values 549 current across 116 DC analysis 113 de-energizing 115 electromotive force 116 energizing 114 exercises 120 out of sync 119 parallel 112 series 112 variable 111 instrumentation and bench 509 integral 15 integrator (operational amplifiers) 505 International System of Units (SI) 521

565

566

Index

inverting (operational amplifiers) 488 inverting polarities 30 inverting summing (operational amplifiers–multiple inputs) 502

k Kirchhoff’s laws 197, 216, 219 example 1 199 example 2 205 exercises 210 first 198 second 199

l lead-free solder 519 light bulb 133 limits 10

m magnetic field, connection 123 mathematical concepts xxvi meshes 197, 200 metric system 539 Millman’s theorem 287 admittance 290 example 1 291 exercises 294 theory 287 mountain analogy 32 multimeter 509 root mean square (RMS) 509

n nodal analysis 215 example 1 215 example 2 219 example 3 221 exercises 226 nodes or junctions 197 non-inverting (operational amplifiers) 488

non-inverting summing (operational amplifiers) 2 inputs 500 3 inputs 501 multiple inputs 497 Norton’s theorem 257 current source 258 equivalent circuit 258 example 258 exercises 263 methodology 258 NPN junction 143

o ohmmeter 512 Ohm’s laws 51 example 1 53 example 2 53 example 3 55 example 4 56 example 5 57 exercises 58, 59 first law 52 second law 53 operational amplifiers 485 characteristics 488 differentiator 507 how they work 486 integrator 505 inverting 488 summing (multiple inputs) non-inverting 488 non-inverting summing 2 inputs 500 3 inputs 501 multiple inputs 497 voltage follower 493 oscilloscope 513 ovens 184

p parallel impedances 166 parallel inductance 112

502

Index

phase 88, 165, 177 phase angle 325 polar numbers 523 positive cycle 463 potentiometer 44 power exercises 191 factor 190, 325 in parallel 182 in series 183 and work 181 powers of 10 1 adding and subtracting 4 computers and programming 6 division 4 engineering notation 6 expressing roots as exponents 5 extracting root 5 multiplication 3 operations 3 raising to number 4 roots 2 power supply 517 rails 452

q quiescent operating point

453

r RC circuits 297 apparent power 326 charge during discharge 311 charge equation 301 charging a capacitor 297 charging current 305 charging voltage 298 current during discharge 314 discharging a capacitor 310 example 1 315 example 2 317 example 3 320 example 4 322 exercises 328

general formula (time constant) 309 impedance 323 phase angle 325 power factor 325 reactive power 326 real power 326 root mean square (RMS) current 325 time constant 308 transient and steady states 308 transient phase duration 309 voltage during discharge 314 reactive power 189, 326 real numbers 529 real power 326 relative voltages 31 resistance 41 Thévenin 236 resistors 41 color code 42, 523 commercial values 541 DC and AC analysis 46 electric characteristics 45 exercises 48 important facts 45 input and output synchronism 47 in parallel 46 power 42 practical usage 45 in series 45 types 42 reverse biasing 138 right-hand rule 105 RLC circuits 377, 427 analysis 380 critically damped solution current 430 current curve 431 voltage 388 voltage curve 396 current 427, 430 example 1 408, 432 example 2 437

567

568

Index

RLC circuits (cont’d) exercises 418, 442 overdamped solution current 431 current curve 431 voltage 396 voltage curve 397 series 377 transient solution 384 underdamped solution current 431 current curve 432 voltage 401 voltage curve 408 voltage across the capacitor 380 RL circuits 341 de-energizing 349 current during 349 voltage during 352 energizing 341 current during 342 voltage during 346 example 1 354 example 2 355 example 3 358 exercises 362 time constant 353 transient and steady states 353 root mean square (RMS) current 325 multimeter 509 value 186, 525

s scientific method 1 scientific notation as a tool second oven 185 semiconductors 133, 135 barrier voltage 140 biasing curve 138 bipolar junction 135

2

current and voltage 141 forward biasing 138 light bulb 133 negative structures 135 positive structures 137 pure magic 137 reverse biasing 138 thermal voltage 139 series impedance 166 series inductance 112 simulation tools 557 sinusoidal function 185 soldering fume extractors 517 soldering station 517 source transformations 151 example 153 exercises 160, 161 stored energy 79 strong force and weak force 21 superposition theorem 269 example 270 exercises 281 methodology 269 symbols 41

t table of integrals 537 terminology 27 thermal runaway 459 thermal voltage 139 Thévenin’s theorem 235 circuit, equivalence 235 equivalent 249 behavior 245 resistance 236 voltage 240 example 236 exercises 250 methodology 236 voltage 243 transformers 123

Index

center tap 125 direct current 127 internal resistance 127 law of conservation of energy 126 leakage flux 127 multiple secondaries 125 primary and secondary 124 real-life 125 transistor 144, 451, 452 AC analysis 462 amplifier design 471 biasing 455 class A amplifier 454 collector current 459, 460 collector feedback biasing 466 common emitter 455 exercise 477 fixed base biasing 455 hFE 452 how it works? 145 positive cycle 463 setting it up 458 thermal runaway 459 voltage divider biasing 469 trigonometric numbers 535 trimpots 44

v variable capacitors 93 variable inductor 111 very large numbers 2 very small numbers 3 virtual ground 488 voltage and current 81 exercises 38, 39 sources 147 voltage divider biasing 469 voltage follower (operational amplifiers) 493 voltage on charging capacitor 82 voltage sources, dependent 148 voltage, Thévenin 240 voltmeter 510

w water storage 451 wires American wire gauges 539 cross sections 539 diameter 515 metric system 539 types 516 Wye–Delta conversion 65

569